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Starting from rest at the bottom, a crate of mass mm is being continuously pus...
May 28, 2024
Solution
a
Draw a Free Body Diagram (FBD) for the crate. The forces acting on the crate are: - The gravitational force mgmg acting vertically downward. - The normal force NN acting perpendicular to the slope. - The horizontal force FF acting horizontally to the right. - The component of the gravitational force parallel to the slope, mgsin⁡θmg \sin \theta. - The component of the gravitational force perpendicular to the slope, mgcos⁡θmg \cos \theta
b
Find the distance dd that corresponds to the height hh. Using trigonometry, we have: sin⁡θ=hd  ⟹  d=hsin⁡θ\sin \theta = \frac{h}{d} \implies d = \frac{h}{\sin \theta}
c
Review: Find the normal force. The normal force NN is given by balancing the perpendicular forces: N=mgcos⁡θN = mg \cos \theta
d
Find the work done by each force acting on the crate as it moves the distance dd up the incline. - Work done by the horizontal force FF: WF=Fdcos⁡θ=Fhsin⁡θcos⁡θW_F = Fd \cos \theta = F \frac{h}{\sin \theta} \cos \theta - Work done by the gravitational force: Wmg=−mghW_{mg} = -mg h - Work done by the normal force NN: WN=0W_N = 0 (since the normal force is perpendicular to the displacement)
e
Use the work-kinetic energy theorem to find the speed vv of the crate at the final point B. The work-kinetic energy theorem states: Wtotal=ΔKE=12mv2−0W_{\text{total}} = \Delta KE = \frac{1}{2}mv^2 - 0 The total work done is: Wtotal=WF+Wmg=Fhsin⁡θcos⁡θ−mghW_{\text{total}} = W_F + W_{mg} = F \frac{h}{\sin \theta} \cos \theta - mg h Thus, 12mv2=Fhsin⁡θcos⁡θ−mgh  ⟹  v=2m(Fhsin⁡θcos⁡θ−mgh)\frac{1}{2}mv^2 = F \frac{h}{\sin \theta} \cos \theta - mg h \implies v = \sqrt{\frac{2}{m} \left( F \frac{h}{\sin \theta} \cos \theta - mg h \right)}
f
Use kinematics to find the acceleration of the crate. Verify that you get the same answer using Newton's 2nd2^{\text{nd}} Law. - Using kinematics: v2=u2+2ad  ⟹  a=v22d=v2sin⁡θ2hv^2 = u^2 + 2ad \implies a = \frac{v^2}{2d} = \frac{v^2 \sin \theta}{2h} - Using Newton's 2nd2^{\text{nd}} Law: Fnet=ma  ⟹  Fcos⁡θ−mgsin⁡θ=ma  ⟹  a=Fcos⁡θ−mgsin⁡θmF_{\text{net}} = ma \implies F \cos \theta - mg \sin \theta = ma \implies a = \frac{F \cos \theta - mg \sin \theta}{m}
Answer
The final speed vv of the crate at point B is given by: v=2m(Fhsin⁡θcos⁡θ−mgh)v = \sqrt{\frac{2}{m} \left( F \frac{h}{\sin \theta} \cos \theta - mg h \right)}
Key Concept
Work-Energy Theorem: The work done on an object equals the change in its kinetic energy. Equation: W=ΔKEW = \Delta KE (Work equals change in kinetic energy)
Explanation
By calculating the total work done on the crate and using the work-energy theorem, we can determine the final speed of the crate.
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