A mass m = 13 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 13 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4381 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is k = 0.43. The mass leaves the spring at a speed v = 2.9 m/s.

1)How much work is done by the spring as it accelerates the mass?

2)How far was the spring stretched from its unstreched length?

3)The mass is measured to leave the rough spot with a final speed vf = 1.2 m/s.

How much work is done by friction as the mass crosses the rough spot?

4)What is the length of the rough spot?

5)In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?

6)In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

7)Return to a scenario where the blcok makes it throgh the entire rough patch. If the rough patch is lengthened to a distance of three times longer, as the block slides through the entire distance of the rough patch, the magnitude of the work done by the force of friction is:

A.)the same

B.)three times greater

C.)three times less

D.)nine times greater

E.)nine times less

Explanation / Answer

Use conservation of energy to determine the energy of the block as it leaves the spring. The potential energy of the spring becomes the kinetic energy of the block.

As for a spring, F=kx, where F is force, k is the spring constant, and x is the distance the spring is stressed or compressed from equilibrium, U (potential energy) = .5k(x^2), as energy is the integral of velocity with respect to distance (x). K is kinetic energy.

K=U=.5k*x^2
K=.5*(4381 N/m)*(X)^2 = -------- J

Since X1 anbd X2 values are not given cannot caluclate .

We can use this energy value to determine how far the block will go as the friction will "consume" all of this kinetic energy.

Now as you know, the magnitude of the force of kinetic friction on a flat surface is equivalent to...

F=mg

Now the integral of force with respect to displacement (or force * displacement) is work, so...

W=mgy

Where y is your answer...so if we set K=W...

K=mgy
K/(mg)=y
(-------J)/((0.43)*(13 kg)*(9.8 m/s^2)) = y = --------------- m

Question is not complete...... Provide X1 and X2 (distance) values.

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