Friction and Potential Energy in Incline Block Experiment

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A physics student analyzes the motion of a block on an incline, calculating the work done by pulling it up the incline to be approximately 877.85 J. After the block slides down a frictionless surface and compresses a spring, the student struggles to determine the block's speed upon reaching the horizontal surface, initially calculating it as 8.38 m/s, which is incorrect. The discussion highlights the complexities of applying work-energy principles, particularly in relation to kinetic energy and friction. Additionally, the student seeks to find the spring constant and the distance the block rebounds up the incline. The conversation emphasizes the need for a clearer understanding of energy transformations in this scenario.
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A physics student pulls a block of mass m = 25 kg up an incline at a slow constant velocity for a distance of d = 4.5 m. The incline makes an angle q = 33° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.3.

a) What is the work Wm done by the student?
I got this question right. It's 877.8548265 J.

At the top of the incline, the string by which she was pulling the block breaks. The block, which was at rest, slides down a distance d = 4.5 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall. The block hits the spring, compresses it a distance L = 0.8 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline.

b) What is the speed v of the block when it first reaches the horizontal surface?

I don't know how to do this. I thought it would be W=KEf-KEi. So I did 877.8548265 = (.5 x 25 x v2)- (0). I got 8.38 m/s, but its not right.

c) What is the spring constant k of the spring?

d) How far up the incline d1 does the block rebound?
 
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wolves5 said:
b) What is the speed v of the block when it first reaches the horizontal surface?

I don't know how to do this. I thought it would be W=KEf-KEi. So I did 877.8548265 = (.5 x 25 x v2)- (0). I got 8.38 m/s, but its not right.

What is the work equal to? Friction and potential energy are not doing the same job as in a).
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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