Question

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 14.0 N.

(a) What acceleration is produced?
(b) How far does the crate travel in 10.0 s?
(c) What is its speed at the end of 10.0 s?

Answer 1

The crate has an acceleration of 0.43 m/s^2. It travels 21.5 m in 10.0 s and has a speed of 4.3 m/s at the end of 10.0 s.

Step-by-step explanation:

To solve this problem, we can use Newton's second law, which states that force is equal to mass times acceleration (F = ma).

(a) We are given that the mass of the crate is 32.5 kg and the net horizontal force acting on it is 14.0 N. Plugging these values into the equation, we get:

F = ma

14.0 N = 32.5 kg * a

a = 14.0 N / 32.5 kg

a = 0.43 m/s^2

So, the acceleration produced is 0.43 m/s^2.

(b) To find the distance traveled by the crate in 10.0 s, we can use the equation of motion: distance = initial velocity * time + (1/2) * acceleration * time^2.

Since the crate starts at rest, the initial velocity is 0 m/s:

distance = 0 * 10.0 s + (1/2) * 0.43 m/s^2 * (10.0 s)^2

distance = 0 + (1/2) * 0.43 m/s^2 * 100.0 s^2

distance = 21.5 m

So, the crate travels 21.5 m in 10.0 s.

(c) To find the speed of the crate at the end of 10.0 s, we can use the equation of motion: final velocity = initial velocity + acceleration * time.

Since the crate starts at rest, the initial velocity is 0 m/s:

final velocity = 0 + 0.43 m/s^2 * 10.0 s

final velocity = 4.3 m/s

So, the speed of the crate at the end of 10.0 s is 4.3 m/s.

Answer 2

given,

mass of crate = 32.5 Kg

horizontal force = 14 N

initial speed, u = 0 m/s

a) acceleration produced = ?

we know,

F = m a

`a = (F)/(m)`

`a = (14)/(32.5)`

a = 0.431 m/s²

b) using equation of motion

t = 10 s

`s = ut + (1)/(2)at^2`

`s = 0 + (1)/(2)* 0.431* 10^2`

s = 21.54 m

c) again using equation of motion

v = u + at

v = 0 + 0.431 x 10

v = 4.31 m/s