Question

A box A of mass 0.8kg rests on a rough horizontal table and is attached to one end of a light inextensible string. The string passes over a smooth pulley fixed at the edge of the table. (see ? below)

The other end of the string is attached to a sphere B of mass 1.2kg, which hangs freely below the pulley. The magnitude of the frictional force between A and the table is FN. The system is released from rest with the string taut. After release, B descends a distance of 0.9m in 0.8s. Modelling A and B as particles, calculate
a) the acceleration of B
b) the tension in the string
c) the value of F
Sphere B is 0.9m above the ground when the system is released. Given that it does not reach the pulley and the frictional force remains constant throughout,
d) find the total distance travelled by A

Answer 1

a) 2.8125 `"ms^-2"`
b) 8.385N
c)6.135N
d) Assuming that the statement "Given that it does not reach the pulley" is talking about box A, `1.23 m`

Explanation:

a)

Use equations of motion formula: `s=ut+\frac{1}{2}at^{2}`

Let s=0.9 (the distance it descends), t=0.8, u=0 (dropped from rest) and a is what you're trying to work out.

you should then get `a=2.8125m/s^2`

b)

Resolve vertically for sphere B, using Newton's 2nd law of motion:
`"F=ma"`

Where `1.2g-"T"=1.2*2.8125`

Let g=9.8, and we get `"T"=8.385N`

c)
Resolve horizontally with the known fact that since it's a smooth pulley, tension is constant, and acceleration is same for both masses, `2.8125"ms"^{-2}`

Thus giving us: `8.385-F=0.8*2.8125`

Rearranging that to get `F=6.135N`

d) Interpret question statement so box A does not reach pulley. When B reaches the ground, A has reached a velocity given by

`V = u + a*t = 0 + 2.8125m/s^2*0.8 s = 2.25 m/s`

That velocity will decay to zero when force F has done work equal to the kinetic energy when tension T suddenly goes to zero. A's initial velocity is `2.25 m/s` and `F=6.135N`.

`"work" = KE_i`

`F*d = 1/2*m*v^2`

`d = (m*v^2)/(2*F)`

`d = (0.8 kg*(2.25 m/s)^2)/(2*6.135N) = 0.33 m`

A slid a distance 0.9 m while B was pulling it, this additional 0.33 m makes the total distance `1.23 m`.