Question

A tow truck exerts a 18300-N force upon a 1210-kg car to drag it out of a mud puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane of motion of the car is horizontal. Determine the time required to drag the car distance of 6.90 meters from its rest position.

Answer 1

@incredebility

Answer 2

pon neglecting air resistance, there are four forces acting upon the object. The up and down forces
balance each other. The acceleration is rightward (or in the direction of the applied force) since the
rightward applied force is greater than the leftward friction force. The horizontal forces can be
summed as vectors in order to determine the net force.
Fnet = •Fx = 18300 N, right - 17900 N, left = 400 N, right
The acceleration of the object can be computed using Newton's second law.
ax = •Fx / m = (400 N, down) / (1210 kg) = 0.3306 m/s/s, right
This acceleration value can be combined with other known kinematic information (vi = 0 m/s, d = 6.90 m)
to determine the time required to drag the car a distance of 6.9 m. The following kinematic equation is
used; substitution and algebra steps are shown.
d = vi • t + 0.5 •a • t2
d = vi • t + 0.5 •a • t2
6.90 m = 0.5 • (0.3306 m/s/s) • t2
6.90 m / (0.5 • 0.3306 m/s/s ) = t2
41.4 = t2
6.46 s = t