Question

A mysterious rocket-propelled object of mass 47.5kg is initially at rest in the middle of the horizontal, frictionless surface of an ice-covered lake. Then a force directed east and with magnitude F(t)= (17.0N/s )t is applied.
How far does the object travel in the first 5.25s after the force is applied?

Answer 1

Answer:

Distance d=8.63m

Explanation:

mass of object m=47.5 kg

Initial velocity vi=0 m/s

Magnitude of force on object F(t)=(17.0 N/s)t

To find

Distance

Solution

The rocket accelerates due to variable force so we apply Newtons second law but the acceleration will not be constant because the force is not constant

We use  ax=Fx/m to find acceleration but it must be integrate to find velocity and then the distance that rocket travels

So

$ax=Fx/m\\gives\\ax(t)=(17.0N/s)t/(47.5kg)\\ax(t)=(0.35789m/s^{3})t\\ As\\Vx(t)=V_{o}+\int\limits^{}_{} { \lim_{0 \to \ t} ax(t) } \, dt=(0.178945m/s^{3} )t^{2}\\ X-X_{o}=\int\limits^{}_{} { \lim_{0 \to \ t} v(t) } \, dt=(0.05964m/s^{3})t^{3}\\ as\\t=5.25s\\So\\X-X_{o}=(0.05964m/s^{3})(5.25s)^{3}\\X-X_{o}=8.63m$

Distance d=8.63m