If you need 40.0 Nm of torque in order to loosen a nut on a wn
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Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W = ×
W = ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = 0°
W = ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
= ΔKE
= mv² -
mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
=
mv² -
mu²
we make F, the subject of formula
F = ( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = ( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N