A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H =

Question

A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the
ground. It slides down the plane, across a frictionless horizontal floor, and then around a
frictionless loop-the-loop of radius R = 2.0 m. On the floor the speed of the block is observed to
be 11 m/s.
What is the work done by friction on the block as it slides down the inclined plane?

A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop-the-loop of radius R = 2.0 m. On the floor the speed of the block is observed to be 11 m/s. What is the work done by friction on the block as it slides down the inclined plane? What is the magnitude of the normal force exerted on the block at the top of the loop?

Explanation / Answer

Block on the loop

First, calculate the speed at the top of the loop using energy
.5*m*11^2=.5*m*v^2+m*g*2*R
solve for v^2
v^2=11^2-8*9.81
v^2=42.52 m/s

At the top of the loop the normal force is centripetal minus gravity

N=1.8*(42.52/2-9.81)
20.6 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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