Question #127543

A 50.0 kg crate is at the top of a ramp that makes a 30o angle with level ground. If the crate starts to slide and the coefficient of kinetic friction between the crate and the ramp is 0.42, what is the acceleration of the crate?


Expert's answer

As per the question,

Mass of the crate (m)=50kg(m)= 50 kg

Angle of the ramp with the ground level (θ)=30(\theta)=30^\circ

Coefficient of the kinetic friction (μk)=0.42(\mu_k) = 0.42

Acceleration of the crate (a)=?(a) = ?

Now,

ma=mgsinθfs\Rightarrow ma=mg\sin\theta -f_s

Now, substituting the values,

a=mgsinθμkmgcosθm=mg(sin300.42×cos30)ma=\frac{mg\sin\theta-\mu_k mg\cos\theta}{m}=\frac{mg(\sin 30^\circ -0.42\times \cos30^\circ)}{m}

a=9.8×0.3=2.94m/sec2a=9.8\times 0.3 =2.94 m/sec^2


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