Change in Kinetic Energy as Block moves up Incline and starts from rest

[cll123]

Homework Statement

Two blocks, A and B (with mass 20.0 kg and 40.0 kg , respectively), are connected by a string, as shown in Figure P5.64. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between block A and the incline is μk = 0.290. Determine the change in the kinetic energy of the system as block A moves from to C to D, a distance of 19.0 m up the incline if the system starts from rest. (The angle above the horizontal is 37 degrees)

Homework Equations

The Attempt at a Solution

 

[amd123]

Can you show us a picture of the diagram?

 

[nlightner]

As the block A moves up the incline, it gains potential energy due to its increase in height. This gain in potential energy is accompanied by a decrease in kinetic energy, as the block's speed decreases due to the force of friction acting against it. Therefore, the change in kinetic energy of the system is equal to the negative of the change in potential energy, or ΔKE = -ΔPE.

To calculate the change in potential energy, we can use the equation ΔPE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the change in height. In this case, the change in height is equal to the distance the block moves up the incline, which is 19.0 m.

Next, we can calculate the change in kinetic energy using the equation ΔKE = 1/2mv^2, where m is the mass of the block and v is the final velocity. Since the system starts from rest, the final velocity is equal to zero. Therefore, the change in kinetic energy is also equal to zero.

Thus, the change in kinetic energy of the system as block A moves from C to D is equal to the negative of the change in potential energy, or ΔKE = -ΔPE = -(mgh) = -(20.0 kg)(9.8 m/s^2)(19.0 m) = -3724 J. This means that the system loses 3724 J of kinetic energy as block A moves up the incline from C to D.