a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that force?

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AL2006 AL2006 · Answer 1 · 2018-10-04T18:56:30+02:00

Final momentum of the box = (mass) x (speed) = (10kg) x (25m/s) = 250 kg-m/s .

Impulse delivered by the force = final momentum of the box = (force) x (time)

250 kg-m/s = (force) x (4 sec)

Divide each side by (4 sec):

Force = (250 kg-m/s) / (4 sec)

Force = (250/4) kg-m/s²

Force = 62.5 Newtons

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pbsindhuja pbsindhuja · Answer 2 · 2018-10-09T20:49:29+02:00

Given:

mass of the box (m):10 Kg

u(initial velocity)=0.

v(final velocity)=25m/s

t(time taken)=4 sec

Now we know that

v=u+at

Where v is the final velocity which is measured in m/sec

u is the initial velocity which is measured in m/sec

a is the acceleration which is measured in m/s^2

t is the time measured in sec

Substituting the given values in the above formula we get

25=0+ax4

a=6.25 m/s^2

Now we also know that

F= mxa

Where F is the force applied on the object which is measured in N.

m is the mass of the object measured in Kg

a is the acceleration measured in m/s^2.

Now substituting the values in the above formula we get

F=mxa

F=10x6.25=62.5 N