Ms. PB is pushing Mr. Rigney in a wheelchair with a force of 10 N East, while Mr. Rigney is using his arms to
1 answer:
The net force is 6 N East
Explanation:
First of all, we start by noticing that all the forces act along the direction East-West, so we can simply find the net force by using algebraic addition.
In order to find the net force on the wheelchair, we need to define a positive direction and write down all the force with the proper sign.
Let's choose East as positive direction. Therefore, we have the following forces:
(east), the force applied by Ms. PB
(east), the force applied by Mr. Rigney
(west), the frictional force, acting in the opposite direction
(west), the air resistance, acting in the opposite direction
Taking into account the correct signs, we can now find the net force on the wheelchair:
And the positive sign tells us that the direction of the net force is East.
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Answer:
The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.
Explanation:
Given:
Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]
Speed of the cart (v) = 3.0 m/s
Thrust force by the rocket engine (F) = 8.5 N
Vertical height of the loop (y) = 20 m
Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.
Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.
So, we will apply kinematics of motion in the two directions separately.
Vertical motion:
Given:
Force acting in the vertical direction is given as:
So, acceleration in the vertical direction is given as:
Acceleration = Force ÷ mass
Vertical displacement of rocket is same as the height of loop. So,
There is no initial velocity in the vertical direction. So,
Now, applying equation of motion in vertical direction. we have:
Now, time taken to reach the loop is 2.15 s.
Horizontal motion:
There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.
Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,
Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.