Question

A block of mass 2.25 kg starts out at rest at the top of a 15.0 m long plane inclined 30 degrees with the horizontal. (a) What is its initial gravitational potential energy? As it slides down the plane, the block encounters a frictional force of 3.5 N. (b) How much work is done to overcome friction? (c) What is the kinetic energy of the block when it reaches the bottom of the incline? At a distance of 0.5 m from the bottom of the inclined plane is a horizontal spring force constant 52 N/m. Suppose that the coefficient of kinetic friction between the block and the horizontal surface is 0.15. (d) What is the kinetic energy of the block just before it comes in contact with the spring? (e) By how much will the spring be compressed? (f) What is the work done against friction as the block moves along the horizontal surface?

Answer 1

Explanations & Calculations

a)

$\qquad\qquad \begin{aligned} \small E_p&=\small mg\times L\sin\theta\\ &=\small 2.25\times9.8\times15\sin30\\ &=\small 165.4\,J \end{aligned}$

b)

$\qquad\qquad \begin{aligned} \small Work&=\small fL\\ &=\small 3.5N\times15m\\ &=\small 52.5\,J \end{aligned}$

c)

$\qquad\qquad \begin{aligned} \small KE_{at\,bottom}&=\small 165.4\,J-52.5\,J\\ &=\small 112.9\,J \end{aligned}$

d)

$\qquad\qquad \begin{aligned} \small KE_{at\,the\,level\,plane}&=\small 112.9\,J-(fs)\\ &=\small 112.9-(\mu R\times s)\\ &=\small 112.9-(0.15\times[2.25\times9.8]\times0.5)\\ &=\small 111.2\,J \end{aligned}$

e)

$\qquad\qquad \begin{aligned} \small EP_{spring}+fx&=\small 111.2\\ \small \frac{1}{2}\times 52x^2+(0.15\times2.25\times9.8)x&=\small 111.2\\ \small 25x^2+3.3x-111.2&=\small 0\\ \small x&=\begin{cases} &=\small 2.0\\ &=\small -2.1 \end{cases} \end{aligned}$

• Therefore, the compression is $\small 2.0\,m$

f)

$\qquad\qquad \begin{aligned} \small Work_{total}&=\small (0.15\times2.25\times9.8)\times2m+(112.9-111.2)\\ &=\small 8.3\,J \end{aligned}$