a block of wood is at rest on a wooden ramp. when wheels are attached to the block of wood, it moves down the ramp.
203328
without wheels ,frition becomes a greater factor,downforce and down hill force are still always there
491199
If the rock is not being slowly pushed into the ground , then the net force is 0 because the force down is equal to the normal force upwards exerted by the ground.\n\nif the car is really heavy and pushes the rock into the ground, then the net force is greater than 0, in a downwards direction.
14795
A wood chuck could chuch lots of wood if a wood chuck could chuck wood.
346741
The total energy of the hoop as it enters the ramp is the sum of its forward kinetic energy and its rotational energy.  If it is rolling at w rad/sec, it is moving forward at a speed of 2*π*R*w.  The total kinetic energy is then .5*M*[2*π*R*w]^2 + .5*I*w^2   I is the moment of inertia of the hoop, M*R^2.\n\nThe total kinetic energy is converted to potential energy when the hoop stops.  Its potential energy is M*g*h, where h is the height of the hoop above the horizontal portion.  If the hoop travels a distance L up the ramp, it climbs an amount = L*sinø.  Now equating the potential energy at the top to kinetic energy at the bottom you get\n\n.5*M*[2*π*R*w]^2 + .5*M*R^2*w^2 = M*g*L*sinø\n\nYou have all the data to solve for L
489200
You should only ask one question at a time.\n\nI will answer the first:\n\nConsider the forces on the box:\n\nGravity vertically downward\nfriction pulling backwards\nThe force of gravity parallel to the ramp \nminus the force of friction\n=m*a=m*1.22\nso\nm*sin(34)*g-f=m*1.22\nfriction can be expressed as\nf=m*u*cos(34)*g\nnote that mass divides out\nsin(34)*g-u*cos(34)*g=1.22\n\nu=.52\n\nj
438516
Hmmm, 4 ways. You mentioned heat and sound. Good. Also, don't know what the target is made of, let me assume wood - then breaking wood fibers in the target consumed some energy. One more.... (this might seem like grasping) arrows sometime vibrate after hitting a target. That vibration would be where some energy went.
539168
If i understand your scenario, the force is pushing downward on the block.  The horizontal component of the force is 15 N times cosine of 40 degrees.  The vertical component of the force is 15 N times sine of 40 degrees.  However, there is also the force of gravity, which is (3.5 kg * 9.8 m/s2)  (units are N, like the other forces).  So if you add the force of gravity to the vertical component, you have the total downward force.  Multiply this by the coeficient of friction, and you have the force of friction.  \n\nThe force of friction is directed against the direction of motion.  Subtract the frictional force from the horizontal component, and you have the total horizontal force.  Divide the horizontal force by the mass and you should get the acceleration.\n\nfriction force = [(15*sine 40)+(3.5*9.8)]*.25\nacceleration = [(15*cosine 40)-friction force]/3.5
187511
it was about objects in motion
416477
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
366836
as much wood as he could
198344
A fixed axle stays stationary and the tire or wheel spins on it. A live is driven and spins the tire or wheel which is attached with splines that force it to turn with the axle.
172179
If an object is at rest on a surface and you push against it and it doesn't move, is there friction on the object? \nThe answer is "yes", but this kind of friction is different from kinetic friction. Whereas kinetic friction acts to resist the motion of an object sliding across a surface, static friction is the force which keeps a motionless object from being pushed or pulled across a surface.\nIf a wooden block is at rest on a horizontal wooden surface, it is acted upon only by the normal force and the gravitational force.\nIf the surface is inclined by a small angle, θ, a component of the gravitational force acts downward along the surface of the board. The magnitude of this component is mg sinθ. If the block doesn't slide, then it is acted on by the static frictional force, fs, which exactly balances the mg sinθ component of the object's weight.\nIf the inclination of the surface is increased further, the static friction reaches a maximum strength. Unlike kinetic friction, which is roughly constant at low speeds, static friction varies to resist other forces on the block. The formula for static friction, therefore, gives only the maximum possible value of the static frictional force. The formula is as follows:  \n   f (s,mas)= μ × N\nwhere μ is the coefficient of static friction, N is the force normal to the contact surface, and f is the maximum value of the static frictional force This force is exerted in the direction opposite the object's motion. \nIf the surface is inclined so much that the mg sinθ component of the object's weight exceeds fs max, then static friction is overcome, and the block begins to slide. At the angle where the block is just on the verge of slipping, fs max is equal to mg sinθ., and Newton's laws give us that: \nthe sum of all forces in the x-direction is zero\n               μ × N = mgsin θ\nand the sum of all forces in the y-direction is zero\n               N = mgcos θ\nFrom the two equations we find that μ = tan θ\nSo the coeffiecient of static friction is equal to the tangent of the angle of inclination where the block is just on the verge of slipping. \nSo,you need the mass and force but aceleration....check the definition
117049
a lever, a wheel, a pulley...
532883
The question implies you are studying torque.  In part a, calculate the torque about the right sawhorse as a function of the position of the painter.  Find where that torque becomes zero.  For part b, do the same but calculate the torque about the left sawhorse.  You want the points where the plank assembly exactly balances on one sawhorse.
553884
For 4 a&b)\nThere would be three arrows.  The first would be a force in the upward direction (i.e. normal force) second would be force due to gravity(downward direction, with equal magnitude o that of the normal force).  The third arrow is an arrow that is in the same direction as the force applied to the large box.  Its magnitude is proportional to the normal force.  It s static friction that is applying that force.   \n\nAs far as part c goes It's been a while for me.  You've got the textbook, look up some formulas.  Besides, I think you'd need to include some masses in order for me to solve those.\n\nThe force diagram for the Large box is very similar to that of the small box, except that it has forces applied to it from the small box. (this is because the large box is supplying the normal force for the small box.)
364901
The force pulling the block down is m*g*sin(ø);  the force of friction is k*m*g*cos(ø),  The net force is the m*g*sin(ø) - k*m*g*cos(ø).  The resulting acceleration is this force divided by m, so\n\na = g*sin(ø) - k*g*cos(ø)\n\na = g[sin(ø) - k*cos(ø)]
653564
it is a type of Machinery.