Nobel and Maaham are wearing inline skates. Nobel has a mass of 62kg and pushes Maaham, whose

Question

mass is 54kg. Maaham accelerates at 1.2m/s2
[left]. Assume that no friction acts on either person.
a) Determine the force that Nobel exerts on Maaham.
b) Determine Nobels acceleration.

Answer 1

a) To determine the force that Nobel exerts on Maaham, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).

F = m * a

Here, Nobel is pushing Maaham, so the force exerted by Nobel on Maaham is the force required to accelerate Maaham.

So, the force that Nobel exerts on Maaham is:

F = 54 kg * 1.2 m/s^2
= 64.8 N

Therefore, Nobel exerts a force of 64.8 N on Maaham.

b) To determine Nobel's acceleration, we can use Newton's second law of motion again. In this case, Nobel is being pushed back by the reaction force from pushing Maaham. Since there is no friction acting on either person, the reaction force will be equal in magnitude and opposite in direction to the force exerted on Maaham.

So, the force exerted on Nobel by Maaham is also 64.8 N.

Using Newton's second law of motion again:

F = m * a

64.8 N = 62 kg * a

Dividing both sides by 62 kg:

a = 64.8 N / 62 kg
≈ 1.05 m/s^2

Therefore, Nobel's acceleration is approximately 1.05 m/s^2.