Question #169422

A 0.5-kg block, attached to a spring, is pushed until the spring is compressed a distance of 10 cm from the equilibrium position. The block is released from rest and slides along a path that is without friction from point A to point B as shown in Figure 2. After reaching point B, the block travels on a rough ramp of angle 30° from the ground and reaches a height of 1 m (point C). Assuming a spring constant of 5000 N/m and a coefficient of kinetic friction of 0.4.

a. What is the speed of the block at point B?

b. What is the speed of the block at point C?



1
Expert's answer
2021-03-08T08:21:54-0500

a)

KEi+PEi=KEf+PEf,KE_i+PE_i=KE_f+PE_f,0+12kx2=12mv2+0,0+\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+0,v=kx2m=5000 Nm(0.1 m)20.5 kg=10 ms.v=\sqrt{\dfrac{kx^2}{m}}=\sqrt{\dfrac{5000\ \dfrac{N}{m}\cdot(0.1\ m)^2}{0.5\ kg}}=10\ \dfrac{m}{s}.

b)

KEi+PEi+Wext=KEf+PEf,KE_i+PE_i+W_{ext}=KE_f+PE_f,12mvi2+0+(Ffrd)=12mvf2+mgh,\dfrac{1}{2}mv_i^2+0+(-F_{fr}d)=\dfrac{1}{2}mv_f^2+mgh,12mvi2μkmgcosθhsinθ=12mvf2+mgh,\dfrac{1}{2}mv_i^2-\mu_k mgcos\theta \dfrac{h}{sin\theta}=\dfrac{1}{2}mv_f^2+mgh,vf=vi22gh(μkcotθ+1),v_f=\sqrt{v_i^2-2gh(\mu_kcot\theta+1)},vf=(10 ms)229.8 ms21 m(0.4cot30+1)=8.17 ms.v_f=\sqrt{(10\ \dfrac{m}{s})^2-2\cdot9.8\ \dfrac{m}{s^2}\cdot1\ m\cdot(0.4\cdot cot30^{\circ}+1)}=8.17\ \dfrac{m}{s}.

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