Question

Starting from rest on a frictionless floor you move 160 kg crate accelerated by a net force of 60N applied for 4s. Enter units for acceleration as m/s^2.
What is the acceleration of the crate from Newton’s 2nd law?

If it starts from rest, how far does it travel in time of 4 s?

How much work is done by the 60N net force?

What is the velocity of the crate at the end of 4s?

What is the kinetic energy of the crate at this time?

Answer 1

1. Acceleration: $0.375 m/s^2$

Newton's second law states that the net force on an object is equal to the product between the object's mass and its acceleration:

F = ma

where

F is the net force

m is the mass

a is the acceleration

For the crate in this problem we have

F = 60 N

m = 160 kg

Re-arranging the equation, we can find the acceleration of the crate:

$a=\frac{F}{m}=\frac{60 N}{160 kg}=0.375 m/s^2$

2. Distance: 3 m

For this problem we can use the following SUVAT equation:

$S=ut+\frac{1}{2}at^2$

where

S is the distance travelled by the crate

u = 0 is the initial velocity, since it starts from rest

t = 4 s is the time elapsed

$a=0.375 m/s^2$ is the acceleration

Substituting all numbers into the equation,

$S=0+\frac{1}{2}(0.375)(4)^2=3 m$

3. Work done: 180 J

The work done by a force, assuming that the force is parallel to the displacement of the object, is given by

W = F d

where

W is the work done

F is the magnitude of the force

d is the displacement

Here we have

F = 60 N

d = 3 m

So the work done is

$W=(60 N)(3 m)=180 J$

4. Final velocity: 1.5 m/s

The final velocity of the crate can be found by using the following SUVAT equation:

$v^2 = u^2 +2ad$

where here

u = 0 is the initial velocity

$a=0.375 m/s^2$ is the acceleration

d = 3 m is the distance covered

Substituting all the numbers that we have, we found:

$v=\sqrt{u^2 +2ad}=\sqrt{0^2+2(0.375)(3)}=1.5 m/s$

5. Kinetic energy: 180 J

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the velocity

Here we have

m = 160 kg is the mass of the crate

v = 1.5 m/s is its final velocity

So the kinetic energy of the crate is

$K=\frac{1}{2}(160)(1.5)^2=180 J$