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
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answer #1
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answered by sojsail 7
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2016-12-05 07:23:33
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answer #2
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answered by ? 4
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