A 7.9kg carton of coke cans is pushed 7m up a smooth 35° inclined plane by a horizontal force of 65N. if the initial speed of the carton is 4.25m/s up the plane calculate: a. the initial energy of the carton b. the work done by the kinetic energy of the carton c. the work done by gravity d. the work done by normal force e. the final kinetic energy of the block

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

$$\begin{lgathered}F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)\end{lgathered}$$

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

$$\begin{lgathered}0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N\end{lgathered}$$

$$\begin{lgathered}F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N\end{lgathered}$$

$$W_{F_x} = F_xd = 227.5\times 4.8 = 1092J$$

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

$$W_g = -mg\sin(30^\circ)d$$

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

$$W_g = -823J$$

The energy dissipated by the frictional force can be found as follows:

$$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$$

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

$$\begin{lgathered}0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N\end{lgathered}$$

$$W_F = Fd = 227.5\times 4.8 = 1092J$$

$$W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J$$

$$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$$

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

i believe the answer is a

6 cos 34, 6 sin 34 trig