Question

2. A 5.5-kg block is initially at rest on a frictionless horizontal surface. It is pulled with a constant horizontal force of 3.8 N. (a) What is its acceleration? (1) How long must it be pulled before its speed is 5.2 m/s? (c) How far does it move in this time?? ​

Answer 1

Answer:

Although a relatively simple problem, this is another one of those dreaded “2-part” problems that baffle many students.

We will choose one of Newton’s equations of motion shown below:

S=Vit+12at2S=Vit+12at2 ————--equation 1

Vf=Vi+atVf=Vi+at —————equation 2

combine equation 1 and equation 2 to eliminate “t” gives

V2f−V2i=2aSVf2−Vi2=2aS —————equation 3

The problem asks for speed after a given distance, so you would choose either equation 1 or equation 3 since they both have distance ( SS ) in the equation. But since we don’t know the travel time, let’s choose equation 3:

V2f−V2i=2aSVf2−Vi2=2aS

Before we can solve this equation, we first need to calculate the acceleration. This is why I call this a 2-part problem.

F=maF=ma

or

a=Fm=1200N5kg=1200kg⋅ms25kg=240ms2a=Fm=1200N5kg=1200kg⋅ms25kg=240ms2

equation 3:

V2f−V2i=2aSVf2−Vi2=2aS

V2f−0=2(240)(5)Vf2−0=2(240)(5)

Vf=49.0msVf=49.0ms

Answer 2

Explanation:

hope this helps you better