A 12.0kg box initially at rest slides from the top of a ramp. The top of the ramp is 6.0m above the ground. If the force of friction along the 11.0 m long incline is 5.0 N,

Question

A 12.0kg box initially at rest slides from the top of a ramp. The top of the ramp is 6.0m above the ground.If the force of friction along the 11.0 m long incline is 5.0 N, what is the speed of the box as it reaches the bottom of the ramp?

Please show work,

thank you.

Answer 1

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Answer 2

I think the best way to solve this problem is with conservation of energy. We know that the box is initially at rest, and so all the initial energy is in gravitational potential energy. As the box slides down the ramp, the potential energy is converted to kinetic energy, with some loss of energy due to friction. If we can calculate the energy lost to friction and the initial gravitational potential energy, then we can determine the kinetic energy the box has as it reaches the bottom of the ramp, and from that determine its speed.

First, the gravitational potential energy with respect to the bottom of the ramp can be calculated with the equation PE = mgh where m is the mass of the object, g is the acceleration due to gravity, and h is the vertical distance from the reference point (in this case, the bottom of the ramp). We know all of these values, so let's plug them in and calculate the potential energy: PE = (12.0 kg) * (9.81 m/s²) * (6.0 m) = 706.32 J.

Next, we can calculate the work done on the box by friction to determine the energy lost to friction. Work is equal to force multiplied by distance (as long as the force is parallel to the motion, and in this case it is). The friction force is 5.0 N, and the distance over which it is applied is 11.0 m, so the work done by friction (W_f) is (5.0 N) * (11.0) m = 55 J.

Now, if we assume that all the potential energy is converted to kinetic energy except for that which is lost to friction, we can calculate the kinetic energy at the bottom of the ramp: KE = PE - W_f = 706.32 J - 55 J = 651.32 J.

*Note that we should have three significant figures in our answer at this point, because while 55 J only has 2 significant figures, the subtraction rule for significant figures cares about the place of last digit (which for 55 J, is the ones place). So, we should treat up to the ones place of our answer as significant, which is three digits in our answer (651.32).

Finally, we can calculate the speed at which the box is moving. The equation for kinetic energy is KE = (1/2)mv², and solving for velocity, we get v = √(2*KE/m). Plugging in our known mass and known kinetic energy, we get v = √(2 * (651.32 J) / (12.0 kg)) = 10.419 m/s. We should report this to three significant figures as both our kinetic energy and mass have three significant figures, so our final answer is 10.4 m/s.