Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.


f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
The answer is “Impulse acting on it” according to the impulse-momentum theorem.
Answer:
20.18 J
Explanation:
We are given that
Mass of bx=m=1.1 kg
Height=h=1.75 m
Initial speed of box=u=2.99 m/s
Final speed,v=2.56 m/s
We have to find the mechanical energy lost due to friction.
Energy at top=
J
Where 
At bottom,h'=0
Energy at bottom=
Energy lost=Energy(Top)-Energy(bottom)=23.78-3.6=20.18 J