Question

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.

Answer 1

Missing figure and missing details can be found here:
http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....

Solution:
(a) The work done by the spring is given by
$W= \frac{1}{2} k (\Delta x)^2$
where k is the elastic constant of the spring and $\Delta x$ is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
$W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J$

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
$W_W = -F_{//} (x_2 -x_1)$
where  the negative sign is given by the fact that $F_{//}$ points in the opposite direction of the displacement of the cart, and where
$F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N$
therefore, the work done by the weight is
$W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J$