7. Consider the motion of an object given by the position vs. time graph shown.

Question

7. Consider the motion of an object given by the position vs. time graph shown. For what time(s) is the speed of the object greatest? -10 01 23 4567 910 Time (s) (A) At all times from t = 0.0 s t = 2.0 s (D) At all times from t = 5.0 s t-7.0 s (B) At time t-3.0 s (E) At time t=8.5 s (C) At time t = 4.0 s 8. A push broom of mass m is pushed across a rough horizontal floor by a force of magnitude T directed at angle as shown above. The coefficient of friction between the broom and the floor is The frictional force on the broom has magnitude (A) (mg + Tsin® ) (D) (mg-Teo:8) (B) (mg-Tsin® ) (E) mg (C) (mg + Teos8) 3 B 37 A plane 5 meters in length is inclined at an angle of 370, as shown above. A block of weight 20 newtons is placed at the top of the plane and allowed to slide down. The magnitude of the normal force exerted on the block by the plane is most nearly (A) 10 N 9- (B) 12 N (C) 16 N (D) 20 N (E) 33 N 10ON 10. A 100-newton weight is suspended by two cords as shown above. The tension in the slanted cord is (C) 150N (A) 50 N (B) 100N (D) 200 N (E) 250 N

Explanation / Answer

7) speed is greatest where slope is more steeper. At t=4 sec, slope is steeper, hence speed will be greatest at t=4 sec.

8)

frictional force = (coefficient of friction)*(normal reaction on the broom),
in normal case i.e, when you just push the broom
along the horizontal direction, frictional force = N, N= normal reaction,
and as there are only two vertical forces i.e, N and the weight of the body,
they are opposite to each other and the broom does not move in the vertical direction
so as per the second Newton's law of motion, Net force = mass * acceleration= 0
so, mg-N=0,so, N=mg.

But here there is one extra force along the vertical direction
and you get that force by resolving the force T into two components
Tsin(theta)along vertical and Tcos(theta) along horizontal,
The horizontal force Tcos(theta) is responsible for the sliding/movement of broom,
but the vertical force, like N and mg , does not result in any motion along
the vertical direction.
so, net force along the vertical should be = mass*acceleration of broom along vertical
and acceleration along vertical=0,
so, net force along vertical =0,
so, mg-[ N+Tsin(theta)] =0,
so, N = mg - Tsin(theta)
so, frictional force= *N = *(mg-Tsin(theta))

9). F_n = 20*cos(37) = 16 N

10). assuming both cords are 30 degrees then:

100/2 = 50 N

50/sin(30) = 100 N

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