Brain Stem Hub

1) Question : A soccer play forcefully shot a ball. The ball is propelled with an initial velocity of 40 m/s at an angle of 30° above the horizontal. Questions 1-5 are based on this scenario. What are the horizontal and vertical distance does it covers after 3 seconds.

15 m, 100 m

100 m , 15 m

Explanation: Remember to find the horizontal and vertical components of the forces. Remember that the horizontal motions is constant during a projectile.

100 m, 100 m

100m , 15m

Explanation: Remember, to use the vertical and horizontal component of the velocities, and the velocity is constant at the horizontal direction

2) Question : What are the horizontal velocity and vertical velocity of the ball at t=3 seconds?

34 m/s, -9m/s

Explanation: Correct. Remember, speed is constant in the horizontal direction and it changes in the vertical direction. Note that acceleration= -10 m/s^2 in the vertical direction due to gravity.

-9 m/s, 34 m/s

34 m/s, 9 m/s

- 34 m/s, -9m/s

3) Question : How long does it take for the ball to reach its maximum height?

1 second

2 seconds

3 seconds

4 seconds

4) Question : How long it takes the ball to reach the ground?

2 seconds

4 seconds

6 seconds

8 seconds

5) Question : A boy weighing 50 kg is sitting 5 meters away from a fulcrum. How far from the fulcrum should a 25 kg lady sit on the opposite side to balance the fulcrum?

2.5 meters

25 meters

1.0 meter

10 meters

6) Question : Two boys weighing 50 kg each are sitting on the left side of a fulcrum, one 5 meters away and the other 6 meters away. How far on the right side should a 20 kg lady sit to balance the fulcrum?

20 m

27.5 m

32.5 m

37.5 m

7) Question : Shown below is a box with of size 200 kg. Two people are pulling the box as follows: Person A is pulling the box to the right with a force of 50 N and angle 60 degrees. Person B is pulling the box to the left with a force of 100N and angle 30 degrees. Find the net horizontal acting on the object

60 N, to the right

50 N to the right

50 N to the left

-60 N to the left

8) Question : Shown below is a box with of size 200 kg. Two people are pulling the box as follows: Person A is pulling the box to the right with a force of 50 N and angle 60 degrees. Person B is pulling the box to the left with a force of 100N and angle 30 degrees. Find the magnitude of the acceleration and its direction.

0.6 m/s^2 to the right

0.6 m/s^2 to the left

0.3 m/s^2 to the right

0.3 m/s^2 to the left

9) Question : Shown below is a box with of size 200 kg. Two people are pulling the box as follows: Person A is pulling the box to the right with a force of 50 N and angle 60 degrees. Person B is pulling the box to the left with a force of 100N and angle 30 degrees. Find the magnitude of the Normal force?

1866 N upward

Explanation: Remember, Normal force is always perpendicular to the surface and directed upwards. Also, remember to use the vertical components of the force.

1866 N downward

2144 N upward

2144 N downward

10) Question : Shown below is a box with of size 200 kg. Two people are pulling the box as follows: Person A is pulling the box to the right with a force of 50 N and angle 60 degrees. Person B is pulling the box to the left with a force of 100N and angle 30 degrees. The object remains in place. What is the magnitude of the static friction given that the coefficient of static friction is 0.5

633 N

733 N

833 N

933 N

Explanation: Use the formula for the static friction

11) Question : An object weighing 10 kg is sliding on an inclined surface. The surface forms an angle 30 degrees with the horizontal as shown below. Find the direction of the weight of the object.

parallel to the inclined surface

Explanation: No, this is the direction of the horizontal component of the weight.

Perpendicular to the surface, downward

Explanation: No, this is the direction of the vertical component of the weight.

Directly upward

Directly downwards

Explanation: Regardless the position of the object, the weight force is always directly downward. The horizontal component of the weight is to the right, parallel to the inclined surface. The vertical component of the weight is downwards, perpendicular to the inclined surface.

12) Question : An object weighing 10 kg is sliding on an inclined surface. The surface forms an angle 30 degrees with the horizontal as shown below. Find the direction and the magnitude of the normal of the object.

Upward, perpendicular to the inclined surface. 86 N

Explanation: First, the normal force is the force exerted by the surface that is perpendicular to it. On an inclined plane, the normal force is a component of the object's weight, which acts vertically downward. The normal force, it turned out, is equal to the vertical component of the weight. There are two forces acting on the object vertically: Normal force and the vertical component of weight. Since, Fverticalnet= zero: there is no vertical movement of the object, this means the vertical component of the weight(Wy) is equal to normal force. Wy= w cos30= 100N x 0.86= 86N. See the image below on how I drew the forces, and it turned out that the Wy is adjacent to the angle 30 degrees

Downward, perpendicular to the inclined surface. 86 N

Upward, perpendicular to the inclined surface. 100 N

Downward, perpendicular to the inclined surface. 100 N

13) Question : An object weighing 10 kg is sliding on an inclined surface. The surface forms an angle 30 degrees with the horizontal as shown below. Calculate the direction and the magnitude of the horizontal component of the weight

100 N

50 N

25 N

5 N

14) Question : An object weighing 10 kg is sliding on an inclined surface. Calculate the acceleration of the object when it slides downwards, given that there is no friction.

1 m/s^2

5 m/s^2

Explanation: Fnet horizontal is equal to the horizontal component of theweight( the question said assume that there is no friction).

10 m/s^2

15 m/s^2

15) Question : An object weighing 10 kg is sliding on an inclined surface. The object is moving at a constant speed. What is the magnitude of the kinetic friction once the object moves.

50 N

Explanation: When the object moves at a constant speed, the net force along the incline is zero because the force of kinetic friction balances the component of the weight acting parallel to the incline.

100 N

150 N

200 N