Question

A net force of 2.5x10^3 North acts on an object for 5.0 s. During this time an object accelerates from rest to a velocity of 48km/h. What is the mass of the object?

Answer 1

Given information:

`F=2.5*10^3 N=2500 N\\v_0=0\\v_f=48 (km)/(h)`

Convert the units of velocity into something more useful for this problem:

`((48 km)/(1 h)) ((1 h)/(3600 s) )((1000 m)/(1 km) )=(48000 m)/(3600 s) =(40)/(3) (m)/(s)`

We know
`F=ma`. Solving for mass gives
`m=(F)/(a)`

Now use a kinematic equation to find acceleration in terms of known information.

`v=v_0+at\\a=(v-v_0)/(t)`

Plug in and solve.

`m=(F)/(a) =(F)/((v-v_0)/(t))\\m=(2500 N)/(((40)/(3)(m)/(s)  -0(m)/(s) )/(5.0 s))=(1875)/(2) kg=937.5 kg`

This gives us a mass of 937.5 kg.

However, the least amount of significant figures we had was 2, so with correct significant figures, this would be 940 kg.