A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1860 N to the right. What is the acceleration of the crate? (Hint: remember to find net force first)

The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6

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aachen aachen · Answer 2 · 2019-10-04T21:16:24+02:00

Answer:

[tex]6\text{m}/\text{s}^2[/tex]

Explanation:

Given: A [tex]230[/tex] [tex]\text{kg}[/tex] steel crate is being pushed along a cement floor. The force of friction is [tex]480[/tex][tex]\text{N}[/tex] to the left and the applied force is [tex]1860[/tex][tex]\text{N}[/tex] to the right.

To Find: What is the acceleration of the crate.

Solution:

Mass of steel crate[tex]=230\text{N}[/tex]

Force of friction on steel crate [tex]=480\text{N}[/tex]

Force applied on steel crate [tex]=1860\text{N}[/tex]

Now,

As force of friction is in opposite direction of force applied

Net force applied on steel crate [tex]=\text{total applied force}-\text{total friction foce}[/tex]

[tex]1860-480[/tex]

[tex]1380\text{N}[/tex]

[tex]\text{force}=\text{mass}\times\text{acceleration}[/tex]

[tex]\text{acceleration}=\frac{\text{force}}{\text{mass}}[/tex]

[tex]\text{acceleration}=\frac{1380}{230}[/tex]

putting values

[tex]\text{a}=6\text{m}/\text{s}^2[/tex]

Hence acceleration of crate is [tex]6\text{m}/\text{s}^2[/tex]