A small block of mass m, which weighs 12.054 N, is hit so that it slides up a fr

Question

A small block of mass m, which weighs 12.054 N, is hit so that it slides up a frictionless inclined plane. The surface of the inclined plane is at an angle theta relative to the horizontal. Just after the mass is hit it has a spend of 17.70 m/s. The mass slides up the incline and travels a distance D along the incline before starting to slide back down. Find the work done by gravity on the block from its starting position up the ramp over the distance D. (Use g = 9.80 m/s^2.) It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 30. 0 m/s A small block of mass m = 1.7 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point Pat rest a distance h = 54.0 in above the bottom of the loop of radius R = 18.0 m. What is the kinetic energy of the mass at the point A on the loop? What is the magnitude of the acceleration of the mass at the point A of the loop? What is the minimum height h for which the block will reach point A on the loop without leaving the track?

Explanation / Answer

8) gain in kinetic enrgy = loss in potential enrgy

= m*g*h - m*g*2*R

= m*g*(h - 2*R)

= 1.7*9.8*(54 - 2*18)

= 299.88 J

9) 0.5*m*v^2 = KE

v = sqrt(2*Ke/m)

= sqrt(2*299.88/1.7)

= 18.78 m/s

a = v^2/R

= 18.78^2/18

= 19.6 m/s^2

10) minimum speed, v = sqrt(g*R)

v = sqrt(9.8*18)

= 13.28 m/s

let h_min is the minimum height

h_min = 2*R + v_min^2/(2*g)

= 2*18 + 13.28^2/(2*9.8)

= 36 + 9

= 45 m

11) W(gravity) = chnage in kinetic enrgy

= -0.5*m*v^2

= -0.5*(12.054/9.8)*17.7^2

= -192.67 J

12) Apply, v^2-u^2 = 2*a*d

a = (v^2-u^2)/(2*d)

so, u1^2/2*d1 = u2^2/2*d2

==> d2 = d1*(u2/u1)^2

= 41.14*(30/11)^2

= 306 m

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