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A 6.0 kg block is pushed 8.0m up a rough 37° inclined plane by a force of 75N that is parallel to the inclined plane. If the initial speed of the block is 2.2 m/s up the plane and a constant kinetic friction force of 25N opposes the motion, calculate (a) the initial kinetic energy of the block; (b) the work done by the 75N force; (c) the work done by the friction force; (d) the work done by gravity; (e) the work done by the normal force; (f) the final kinetic energy of the block.

1 Answer

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Given

m = 6kg

x = 8m

F = 75 N

vo = 2.2 m/s up

Ff = 25 N

Procedure

(a) Initial kinetic energy of the block

In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.


\begin{gathered} K=(1)/(2)mv^2 \\ K=(1)/(2)6\operatorname{kg}\cdot(2.2m/s)^2 \\ K=14.5\text{ J} \end{gathered}

(b) the work done by the 75N force;


\begin{gathered} W=F\cdot d \\ W=75N\cdot8m \\ W=600\text{ J} \\ \end{gathered}

(c) the work done by the friction force


\begin{gathered} W=F\cdot d \\ W=-25N\cdot8m \\ W=-200J \end{gathered}

(d) the work done by gravity


\begin{gathered} W=\text{mg}\cdot d\cdot\sin (37) \\ W=6\operatorname{kg}\cdot9.8m/s^2\cdot8m\cdot\sin (37) \\ W=-283\text{ J} \end{gathered}

(e) the work done by the normal force

0J, perpendicular force to the motion

(f) the final kinetic energy of the block.​


\begin{gathered} W_T=14.5\text{ J}+600J-200J-283\text{ J} \\ W_T=131.5J \end{gathered}

A 6.0 kg block is pushed 8.0m up a rough 37° inclined plane by a force of 75N that-example-1
User TomHastjarjanto
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