A 2.762 kg block is on a horizontal surface with k = 0.160, and is in contact wi

Question

A 2.762 kg block is on a horizontal surface with k = 0.160, and is in contact with a lightweight spring with a spring constant of 784 N/m which is compressed. Upon release, the spring does 1.004E+1 J of work while returning to its equilibrium position. Calculate the distance the spring was compressed.

The force that acted on the mass comes from the spring and from friction. Thus the amount of work equals the kinetic energy of the spring. From the equation of work done by both forces we can calculate the velocity of the spring.

0.16m is not the answer to the first one and 2.03 m/s is not the answer to the second one. So please right answers and work so I can see what was done

The force that acted on the mass comes from the spring and from friction. Thus the amount of work equals the kinetic energy of the spring. From the equation of work done by both forces we can calculate the velocity of the spring.

0.16m is not the answer to the first one and 2.03 m/s is not the answer to the second one. So please right answers and work so I can see what was done

Explanation / Answer

Let the spring was compressed by distance x
work done by spring + energy lost against friction = spring potential energy
1.004*10^1 + miu*m*g*x = 0.5*k*x^2
10.04 + 0.16*2.762*9.8*x = 0.5*784*x^2
392x^2 - 4.33x - 10.04 = 0
solving above quadratic equation, only positive value of x is
x = 0.166 m
This is the answer

Now use:
final kinetic energy =work done by spring
0.5*m*v^2 = 1.004*10^1
0.5*2.762*v^2 = 10.04
v= 2.7 m/s

This is the answer of 2nd part

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