Question

A force of 40.0 N is applied in the positive x-direction to a block of mass 9.05 kg, at rest on a frictionless surface.

(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 5.00 s?
m/s

Answer 1

[AI-generated answer — Explain Bot]

To find the block's acceleration in part (a), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

The formula for Newton's second law of motion is:

F = m * a

where F is the net force, m is the mass, and a is the acceleration.

In this case, the net force acting on the block is 40.0 N, and the mass of the block is 9.05 kg. Therefore, we can rearrange the formula to solve for the acceleration:

a = F / m

Substituting the given values, we get:

a = 40.0 N / 9.05 kg

Calculating this, we find that the block's acceleration is approximately 4.42 m/s² in the positive x-direction.

For part (b), we need to calculate how fast the block is going after 5.00 s. Since the block starts from rest and moves with a constant acceleration, we can use the kinematic equation:

v = u + a * t

where v is the final velocity, u is the initial velocity (which is 0 m/s since the block starts from rest), a is the acceleration, and t is the time.

Substituting the given values, we have:

v = 0 m/s + 4.42 m/s² * 5.00 s

Simplifying the equation, we find that the block's velocity after 5.00 s is approximately 22.1 m/s in the positive x-direction.