A horizontal force of 200N is applied to a 55-kg cart across a 15-m level surfac

Question

A horizontal force of 200N is applied to a 55-kg cart across a 15-m level surface. If the cart accelerations at 2.0 m/s^2, then what is the work done by the force of friction as it acts to retard the motion of the cart? A 20.0-kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20degree with the horizontal. If there is no friction between ramp surface and crate, what is the kinetic energy of the crate at the bottom of the ramp/(g = 9.8 m/s^2) The theorem can be expressed in terms of the work done by both conservative forces, w_c, abd nonconservative forces, w_nc w_c + w_nc = deltaKE

Explanation / Answer

1j=1newton*meter
force=mass*accel
200N=55x
200/55=3.636
3.636-2=1.636
1.636 is the deceleration resulting from friction hence the force of friction is 1.636*55=90newtons
90newtons*distance of 15 meters= 1350 j of work done by friction

2. Emech is conserved

Emech f =E mech i

Kf=mghi=20x9.81x2sin20

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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