Question

A physics student pulls a block of mass m = 22 kg up an incline at a slow constant velocity for a distance of d = 4.5 m. The incline makes an angle ? = 32° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.3.

1) What is the work Wm done by the student?

2) At the top of the incline, the string by which she was pulling the block breaks. The block, which was at rest, slides down a distance d = 4.5 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall. The block hits the spring, compresses it a distance L = 0.7 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline.

What is the speed v of the block when it first reaches the horizontal surface?
v =

3) What is the spring constant k of the spring?
k =

4) How far up the incline d1 does the block rebound?
d1 =

Answer 1

a)

$\large W=F\times d$

$\large F= \left (sin(32^0)\times 22\times 9.8 \right )+\left (cos(32^0)\times 22\times 9.8\times 0.3 \right )$

$\large F= 169.10N$

and the distance is 4.5 m

$\large W=169.10N\times 4.5m$

$\large W=760.96J$

b)

$\large 4.5\times 22\times 9.8\times (sin(32^0)-cos(32^0)\times 0.3)= \frac{1}{2}\times 22\times v^2$

$\large v=4.93m/s$


c)

since L=0.7 m

$\large \frac{1}{2}\times 22\times (4.93)^2=\frac{1}{2}\times k\times 0.7^2$
solve for k

$\large k=1091N/m$

d) Again with energy

$\large \frac{1}{2}\times 22\times 4.93^2= d\times 22\times 9.8\times (sin(32^0)+cos(32^0)\times 0.3)$

solve for d

$\large d=1.58m$