Question

A 50 kg skater pushed by a friend accelerates 5 m/sec2. How much force did the friend

apply?
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Answer 1

Answer:

250 Newton

Explanation:

according to the formula

F=ma (m=50kg and a=50m/s2)

F=50×5=250 and unite of force =Newton

=250 N

Answer 2

Answer:

The force applied by his friend will be equal to 250N.

Explanation:

We have given , the mass of the skater, m = 50Kg

The acceleration produced , a = 5m/s²

To find the force applied by his friend, we can use Newton's second law of motion.

F = ma                                                                           .................(1)

where m is mass of the body and a is the acceleration produced in the body. If we doubled the acceleration then the force will be doubled as well.

Substitute the value of acceleration and mass in the equation (1);

$F= (50kg)*(5ms^{-2})$

$F= 250\;kgms^{-2}$

$F=250N$

Therefore, the skater's friend pushed him with force equals to 250N.