1. Acceleration: [tex]0.375 m/s^2[/tex]
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:
[tex]a=\frac{F}{m}=\frac{60 N}{160 kg}=0.375 m/s^2[/tex]
2. Distance: 3 m
For this problem we can use the following SUVAT equation:
[tex]S=ut+\frac{1}{2}at^2[/tex]
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
[tex]a=0.375 m/s^2[/tex] is the acceleration
Substituting all numbers into the equation,
[tex]S=0+\frac{1}{2}(0.375)(4)^2=3 m[/tex]
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
[tex]W=(60 N)(3 m)=180 J[/tex]
4. Final velocity: 1.5 m/s
The final velocity of the crate can be found by using the following SUVAT equation:
[tex]v^2 = u^2 +2ad[/tex]
where here
u = 0 is the initial velocity
[tex]a=0.375 m/s^2[/tex] is the acceleration
d = 3 m is the distance covered
Substituting all the numbers that we have, we found:
[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2+2(0.375)(3)}=1.5 m/s[/tex]
5. Kinetic energy: 180 J
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
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
[tex]K=\frac{1}{2}(160)(1.5)^2=180 J[/tex]
