F – to the right F=3700 N
F(fr) to the left F(fr) =189 N
mg – down mg=4700•9.8=46060 N
N – up N=mg=46060 N
F(net) =F-F(fr)=3700-189=3511 N
ma=F(net)
a=F(net)/m=3511/4700= 0.74 m/s²
a ↑if F(net)↑ or a↓
mg (action) N (reaction)
A 4700 kg truck on a level surface experiences a force applied due to the engine of 3700 N forward. If the force of friction acting on the truck is 189 N backward:
a) Determine the values of the other forces acting on the truck.
b) Draw a free body diagram and determine the net force.
c) Determine the acceleration of the truck.
d)What are two ways that the acceleration could be increased?
e) What action reaction force pairs are involved in this scenario?
1 answer