A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260,
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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.