Question

Superman jumped off the ground and accelerated his `120` `kg` frame at a rate of `30` `m``/s^2`. What was the force that was the Superman applied to the ground as he jumped?

Answer 1

`4,800 color(white)(l) "N"`

Explanation:

A free body diagram might help illustrate the situation

The block of mass resembles the Superman. He experiences two concurrent forces: the normal force from the ground pointing upwards `"N"` and the gravitational pull `"W"` in the opposite direction.

`"W"=m*g=120 color(white)(l) "kg" * 10 color(white)(l)"m" * "s"^(-2)=1,200 color(white)(l)"N"`

The question states that the Superman experiences an upward acceleration of `30 color(white)(l) "m" * "s"^(-1)`. Newton's Second Law of Motion states that the magnitude of the net force is proportional to its acceleration. That magnitude would be the same as the product of the object's mass and acceleration if the measurement was in `"SI"` units where `1color(white)(l)= 1color(white)(l)"kg"* 1color(white)(l)"m"*"s"^(-2)`.

`Sigma F = m*a=120 color(white)(l) "kg" * 30 color(white)(l)"m" * "s"^(-2)=3,600 color(white)(l)"N"`

From the free-body diagram, the magnitude of the net force is equal to the difference in the strength of the normal force and the gravitational pull on the object. That is:

`Sigma "F"="N"-"W"`

Rearrange the equation and solve for `"N"`

`"N"=Sigma "F" + "W"=3, 600 + 1, 200= 4, 800 color(white)(l) "N"`

By Newton's Third Law of Motion, the force the Superman exerts on the ground is the opposite as the normal force he receive from the ground. The two forces are the same in magnitude. Therefore, the force he applies to the ground is of `4, 800 color(white)(l) "N"`.