A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor

Question

A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force f equal 130 N The coefficient of friction between box and floor is 0.300 Find (a) the work done by the applied force, (b) the change in the kinetic energy of the box, and (c) the final speed of the box. 3. A sled is initially given a shove up a frictionless 23degree incline It reaches a maximum vertical height 2.0 m higher than where it started at the bottom What was its initial speed?

Explanation / Answer

Work done by applied force=F.S=FS=130N*5=650N.m

Force due to friction =coeff of friction* normal=0.3*N=0.3*40*9.8=117.6N

Net force in horizontal direction=130-117.6=12.4N

acceleration=F/M=12.4/40=0.31m/s2

apply equation of motion

v2=u2+2as

v2=2*0.31*5=3.1

v=1.761m/s

change in kinetic energy=0.5mv2-0.5mu2=0.5mv2 (as u=0)

change in kinetic energy=0.5mv2=0.5*40*1.7612=62Nm

final speed of box=v=1.761m/s

3)as component of acceleration due to gravity along shove downward=gsin23

as final speed =0

v2=u2+2as

u2=-2as= -2*(-9.8sin23)*(2/sin23)

u=4.43m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.