Question

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by an applied force of 160 N and a 20 N force of friction. What acceleration is produced? How far does the crate travel in 10.0 s? What is its speed at the end of 10.0 s?

Answer 1

Given:

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by an applied force of 160 N and a 20 N force of friction.

To find:

• Acceleration produced?
• Distance travelled in 10 sec?
• Speed at end of 10 sec?

Calculation:

Let net acceleration be a :

$\therefore \: a = \dfrac{net \: force}{mass}$

$\implies \: a = \dfrac{160 - 20}{3.5}$

$\implies \: a = 4.3 \: m {s}^{ - 2}$

Let distance travelled be d :

$\therefore \: d = ut + \dfrac{1}{2} a {t}^{2}$

$\implies \: d = 0 + \dfrac{1}{2} \times 4.3 \times {(10)}^{2}$

$\implies \: d = \dfrac{1}{2} \times 4.3 \times {(10)}^{2}$

$\implies \: d =215 \: m$

Let final velocity be v :

$\therefore \: v = u + at$

$\implies \: v = 0 + (4.3 \times 10)$

$\implies \: v = 43 \: m {s}^{ - 1}$

Hope It Helps.

Answer 2

Answer: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.