Please help in solving this question

Question

A force of 35.0 N is applied in the positive x-direction to a block of mass 7.40 kg, at rest on a frictionless surface.

HINT

(a)

What is the block's acceleration (in m/s²)? (Indicate the direction with the sign of your answer.)

 m/s²

(b)

How fast is it going (in m/s) after 4.50 s?

 m/s

Answer 1

Hello Jessica;

We are using Newton's 2^nd Law to find the acceleration first.

F = ma ==> 35 = 7.40 x a

a = +4.7 m/^s in the +x-direction.

b) Acceleration is the rate of change in velocity;

V(t) = V₀ + at, (we insert t = 4.5s into the equation and find the velocity at that time)

V(4.5) = 0 + 4.7 x 4.5 ==> Velocity at 4.5s is 21.15 m/s ≈ 21.2 m/s (+x-direction)