A mass and pulley system as shown is released from rest on a frictionless surfac

Question

A mass and pulley system as shown is released from rest on a frictionless surface. The masses

are identical. The masses travel through a distance h before the hanging mass lands on a spring

of spring force constant k. The spring compresses by an amount x when it stops the hanging

mass.

(a) How fast is the block on the surface traveling when the hanging mass is stopped. Write it in

terms of the parameters given in the diagram?

(b) What is the compression of the spring when the hanging mass is stopped? Write it in terms of

the parameters given in the diagram?

Evaluate both answers using the following values for the parameters. m = 1 kg, g = 10 m/s^2, h = 1 m, k = 240 N/m?

Points are awarded based on how easy it is to follow what you are doing.

Explanation / Answer

First of all

when the block stops the net force acting the block becomes 0

and the forces acting on it are.. 1) gravitational force in downwards direction (i.e. mg) 2) and spring force in upward direction (i.e. kx)

we have not considered the tension in the string because even after the hanging block stops the mass on the surface keeps moving and the string becomes slack (or in other words tension in the string becomes 0)

so mg - kx = 0

or x = mg/k

now the potenital energy lost by the hanging mass = mg(h+x)

this energy is now stored as potential energy in spring and kinetic energy in the mass on the surface

so equating both

mg(h+x) = 0.5 kx^2 + 0.5mv^2

where v is the velocity of mass on the surface

now we know x = mg/k

substituting this in the above equation

2*mgh + 2*(mg)^2/k = (mg)^2/k + mv^2

2mgh + (mg)^2/k = mv^2

solving above equation for v; we get

-----------------------------------------

v = sqrt(2gh + mg^2/k) |

-----------------------------------------

and we have already calculated

-------------

x = mg/k |

-------------

substituting the values given in the above answers we get

v = 4.52 m/s

x = 41.67 cm

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