A 2 kg block starts at rest and slides down a frictionless ramp that makes an un

Question

A 2 kg block starts at rest and slides down a frictionless ramp that makes an unknown angle with respect to the horizontal floor below. At the bottom of the ramp, after having slid a distance of 10 meters, the speed of the block is observed to be 12m/s. a) What is the angle that the ramp makes with the floor?

b) The block now slides onto the horizontal floor with the same intial speed of 12m/s. The coefficient of kinetic friction between the block and the floor is (mu)k=0.3. From the moment it reaches the floor, how much time does it take the block to stop?

Explanation / Answer

(a) 0.5 mv^2=mgh

0.5mv^2=mg*(10sin(theta))

sin(theta)=0.5*12*12/10*9.8

So ,theta= 47.22 degree

(b)0.5mv^2=umg*x

x=0.5*12*12/0.3*9.8=24.4 meter

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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