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A block with mass M=16kg rests on a frictionless table and is accelerated by a spring with spring constant k=4065N/M after being compressed a distance x1=0.685m from the spring’s unstretched length. The
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A block with mass M=16kg rests on a frictionless table and is accelerated by a spring with spring constant k=4065N/M after being compressed a distance x1=0.685m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance of 2.8m long. For this rough path, the coefficient of kinetic friction is mk=0.46.1) How much work is done by the spring as it accelerates the block?2) What is the speed of the block right after it leaves the spring?3) How much work is done by friction as the block crosses the rough spot?4) What is the speed of the block after it passes the rough spot?5) Instead, the spring is only compressed a distance X2=0.268m before being released. How far in to the rough path does the block slide before coming to rest?6) What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released.

A block with mass M=16kg rests on a frictionless table and is accelerated by a spring-example-1
User Nonin
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\begin{gathered} m_(block)=16\text{ kg} \\ K=\text{ }4065\text{ N/m} \\ x=0.685m\text{ } \\ l=2.8\text{ m} \\ \mu_k=0.46 \\ 1)\text{ To find work} \\ \text{work}=(Kx^2)/(2) \\ \text{work}=\frac{(4065\text{ N/m})(0.685m)^2}{2} \\ \text{work}=953.7\text{ J} \\ \text{The work }done\text{ by the spring is 953.7J} \\ 2)\text{ To find the velocity} \\ \text{Energy 1 = Energy 2} \\ \text{Energy 1 =Energy due to the stretches spring}=(Kx^2)/(2) \\ \text{Energy 2 = Energy due to the block's velocity}=(m_(block)v_(block)^2)/(2) \\ (Kx^2)/(2)=(m_(block)v_(block)^2)/(2) \\ \text{Solving }v_(block) \\ Kx^2=m_(block)v^2_(block) \\ v^2_(block)=(Kx^2)/(m_(block)) \\ v_(block)=\sqrt{(Kx^2)/(m_(block))} \\ v_(block)=\sqrt{\frac{(4065\text{ N/m})(0.685m)^2}{16\text{ kg}}} \\ v_(block)=10.92\text{ m/s} \\ \text{The blocks velocity is }10.92\text{ m/s} \\ \\ 3)\text{ work done by friction} \\ \text{work = Fr}\cdot l \\ Fr=N\cdot\mu_k \\ N=m_(block)\cdot g,g=gravity=9.81m/s^2 \\ N=(16kg)(9.81m/s^2) \\ N\approx157N\text{ } \\ Fr=(157N)(0.46)_{} \\ Fr=72.22N,\text{ then} \\ \text{work = (72.22N)}\cdot(2.8m) \\ \text{work}=202.22J \\ \text{The work }done\text{ by friction is 202.22J} \end{gathered}

User DaveRGP
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