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TiliK225 [7]
3 years ago
14

A 900 N crate is being pulled across a level floor by a force F of 340 N at an angle of 23° above the horizontal. The coefficien

t of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.
Physics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

0.958891203 m/s²

Explanation:

N = Weight of crate = 900 N

\mu = Coefficient of friction = 0.25

Force of friction acting on the force applied

f=\mu N\\\Rightarrow f=0.25\times 900\\\Rightarrow f=225\ N

Force used to pull the crate

F=340cos 23\\\Rightarrow F=312.97165\ N

The net force is

F_n=F-f\\\Rightarrow F_n=312.97167-225\\\Rightarrow F_n=87.97167\ N

Acceleration is given by

a=\frac{F_n}{m}\\\Rightarrow a=\dfrac{87.97167}{\dfrac{900}{9.81}}\\\Rightarrow a=0.958891203\ m/s^2

The magnitude of the acceleration of the crate is 0.958891203 m/s²

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Answer:

Gwen’s assumption of asteroid hit as long term change is incorrect. Asteroid hit is not a long term change, instead, it is a short term change.  

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consider the velocity of the ball towards the wall as negative and away from the wall as positive.

m = mass of the ball = 513 g = 0.513 kg

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