A 10kg block lies to the left of a 5kg block. If the 10kg block is pushed with a force of 30 Newtons at an angle of 30 degrees South of East:
a) determine the net acceleration of the system.
b) find the contact force (magnitude and direction) of the 5kg mass on the 10kg mass.
a) determine the net acceleration of the system.
b) find the contact force (magnitude and direction) of the 5kg mass on the 10kg mass.
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(We have to assume a frictionless surface, since no coefficients are given. It would also have been more consistent and easier to understand the question if it had said 'lies to the west of...' - tell your teacher or textbook to get his/its act together!)
Acceleration is simply a = F/m = 30/15 = 2 m/s/s, and the direction will be in the same direction the force is acting, since there are no other forces acting.
The 5 kg mass is accelerating at 2 m/s/s so the force causing it to accelerate (the force between the blocks) is simply F = ma = 5*2 = 10 N.
This is the force of the 10 kg block on the 5 kg block, accelerating it in the E-30-S direction, but Newton's third law tells us the force of the 5 kg block on the 10 kg block will have the same magnitude and the opposite direction. So its magnitude is 10 N and its direction is 30 degrees north of west.
(a diagram would, as always, be very helpful)
Acceleration is simply a = F/m = 30/15 = 2 m/s/s, and the direction will be in the same direction the force is acting, since there are no other forces acting.
The 5 kg mass is accelerating at 2 m/s/s so the force causing it to accelerate (the force between the blocks) is simply F = ma = 5*2 = 10 N.
This is the force of the 10 kg block on the 5 kg block, accelerating it in the E-30-S direction, but Newton's third law tells us the force of the 5 kg block on the 10 kg block will have the same magnitude and the opposite direction. So its magnitude is 10 N and its direction is 30 degrees north of west.
(a diagram would, as always, be very helpful)