A 1-kg rock that weighs 9.8 N is thrown straight upward at 20 m/s.Neglecting air resistance,the net force that acts on it when it is half way to the top of its path is
A)less than 9.8 N.
B)9)8 N.
C)more than 9.8 N.

Question

Quizplus · Accepted Answer

According to Newton's laws of motion, the net force acting on an object is given by the product of its mass and acceleration. Since the only force acting on the rock is gravity (neglecting air resistance), the net force can be calculated as:

Net force = mass x acceleration

Acceleration due to gravity = 9.8 m/s^2 (direction is downwards)

At the highest point of the trajectory, the velocity of the rock would be zero. Therefore, at the midpoint of the trajectory, the rock would be moving with a velocity of 20/2 = 10 m/s upwards. Using the equation of motion, we can calculate the height reached by the rock at this point:

v^2 = u^2 + 2as

where v = final velocity (0 m/s), u = initial velocity (10 m/s), a = acceleration (-9.8 m/s^2), and s = distance covered.

Solving for s, we get:

s = u^2/(2a) = (10^2)/(2 x (-9.8)) = 5.1 m

At this point, the net force acting on the rock would be:

Net force = mass x acceleration

= 1 kg x (-9.8 m/s^2)

= -9.8 N

Therefore, the best choice is B) 9.8 N, which is the net force acting on the rock at the midpoint of its trajectory. Note that the force is negative because it is acting in the opposite direction to the motion of the rock.

A 1-Kg Rock That Weighs 9