- #1
joker_900
- 64
- 0
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OK there is a slope inclined at an angle z to the horizontal, where z=arcsin0.6. There is a particle of mass m on the slope (A) attached to a string which goes over a pulley at the top corner and to another particle of mass 2m (B)hanging over the side. Coefficient of friction = 0.25.
In the first part i worked out that A accelerates up the slope at 3.92ms^-2
B descends 1m and then the string breaks. Use conservation of energy to find the total distance A moves before coming to rest.
So i thought, use v^2=u^2 + 2as to show that v^2 = 7.84 when the string breaks, and so the KE of A when the string breaks is 3.92m joules. Then as it goes 1m up the slope, it gains 5.88m joules of PE (using trigonometry). So when the string breaks A has 9.8m joules of energy.
Work done against friction = 0.25*mgcosz * x = 1.96mx joules (where x is the distance A travels up the slope after the string snaps)
So surely:
9.8m - 1.96mx = mgh (as at rest, A has only PE)
9.8m - 1.96mx = 0.6xmg
x = 1.25 metres
So total distance = 2.25 metres
But the answer is 1.5 metres!
OK there is a slope inclined at an angle z to the horizontal, where z=arcsin0.6. There is a particle of mass m on the slope (A) attached to a string which goes over a pulley at the top corner and to another particle of mass 2m (B)hanging over the side. Coefficient of friction = 0.25.
In the first part i worked out that A accelerates up the slope at 3.92ms^-2
B descends 1m and then the string breaks. Use conservation of energy to find the total distance A moves before coming to rest.
So i thought, use v^2=u^2 + 2as to show that v^2 = 7.84 when the string breaks, and so the KE of A when the string breaks is 3.92m joules. Then as it goes 1m up the slope, it gains 5.88m joules of PE (using trigonometry). So when the string breaks A has 9.8m joules of energy.
Work done against friction = 0.25*mgcosz * x = 1.96mx joules (where x is the distance A travels up the slope after the string snaps)
So surely:
9.8m - 1.96mx = mgh (as at rest, A has only PE)
9.8m - 1.96mx = 0.6xmg
x = 1.25 metres
So total distance = 2.25 metres
But the answer is 1.5 metres!