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#1) A child holds a sled on a frictionless, snow-covered hill,
inclined at an angle of 30 degree.
If the sled weighs 74N.find the force exerted on the rope by the child. in units of N
What force is exerted on the sled by the hill? units N

#2)There is friction between the block and the table.
The suspended 3kg mass on the left is moving
up, the 3kg mass slides to the right on the table,
and the suspended mass 6 kg on the right is
moving down.
The acceleration of gravity is 9.8m/s^2
What is the magnitude of the acceleration of the system?
in units m/s^2

2007-03-05 13:04:42 · 2 answers · asked by elquida25 1 in Science & Mathematics Physics

2 answers

1) The downhill component of the sled's weight is w*sin30. If the sled is stationary, there is no net force on it, so the child is providing an equal and opposite force.

There is also a component of the weight that is normal to the slope. That's w*cos30. The hill is providing an equal and opposite force.

2) Can't work this without knowing the coefficient of friction. I'll assign a number: 0.1.

Only the middle block has friction with the table.
Ff = mu*Normal = mu*m*g = 0.1*3 kg*9.8 m/s^2
Ff = 2.9 kg*m/s^2 = 2.9 N

The unbalanced 3 kg of the mass on the right provides the force to the right. That force is the weight of 3 kg
W = 3 kg*9.8 m/s^2
The friction inhibits the motion. The net force is
Fnet = W - Ff
The net force then accelerates the 3 masses. Use
F = m*a
where m is the sum of the 3 masses. Solve for a.

2007-03-05 14:54:52 · answer #1 · answered by sojsail 7 · 0 0

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2016-12-05 07:23:33 · answer #2 · answered by ? 4 · 0 0

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