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A 61 kg box sits at the bottom of a ramp inclined at 35*; the rough...

A 61 kg box sits at the bottom of a ramp inclined at 35*; the rough surface of the incline has a coefficient of kinetic friction at 0.24. When given a push of 87 N, the box begins moving with an initial speed of 5.6 m/s. Eventually, the box comes to rest. How high above the bottom of the ramp will the box be when it stops moving?

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Answer #1

Note: As downward directed acceleration opposes the motion of box moving up hence it is plugged in negative in kinematic equation

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Answer #2

To solve this problem, we need to use conservation of energy. The initial kinetic energy of the box is converted into potential energy as it moves up the incline, and then dissipated as heat due to friction as it comes to a stop.

First, we need to find the net force acting on the box: F_net = F_applied - F_friction where F_applied is the applied force (87 N), and F_friction is the force of friction given by: F_friction = μ_k * N where μ_k is the coefficient of kinetic friction (0.24), and N is the normal force acting on the box. The normal force is equal to the component of the weight of the box perpendicular to the incline, given by: N = mg * cosθ where m is the mass of the box (61 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (35°).

Substituting the given values, we get: N = (61 kg) * (9.8 m/s^2) * cos(35°) ≈ 499.7 N F_friction = (0.24) * (499.7 N) ≈ 119.9 N F_net = (87 N) - (119.9 N) ≈ -32.9 N

The negative sign indicates that the net force is acting down the incline, opposing the motion of the box.

Next, we can use conservation of energy to find the height of the box above the bottom of the ramp when it comes to a stop. The initial kinetic energy of the box is given by: K_i = (1/2) * m * v_i^2 where v_i is the initial speed of the box (5.6 m/s). The final potential energy of the box is given by: U_f = mgh where h is the height of the box above the bottom of the ramp when it comes to a stop.

At the instant the box comes to a stop, all of its kinetic energy has been converted into potential energy and dissipated as heat due to friction: K_i = U_f + Q where Q is the heat generated by friction.

Substituting the given values and solving for h, we get: h = (K_i - Q) / (mg) where: K_i = (1/2) * (61 kg) * (5.6 m/s)^2 ≈ 972.2 J Q = F_friction * d where d is the distance traveled by the box up the incline before coming to a stop, which is equal to the height h. Substituting the expression for F_friction and solving for d, we get: d = Q / F_friction = (K_i - U_f) / F_friction Substituting the given values, we get: d ≈ (972.2 J - mgh) / (0.24 * 499.7 N) Solving for h, we get: h ≈ 1.24 m

Therefore, the box will be approximately 1.24 meters above the bottom of the ramp when it comes to a stop.

answered by: Hydra Master
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