what is the answer for this question ? What is the minimum height h for which th

Question

what is the answer for this question ?

What is the minimum height h for which the block will reach point A on the without leaving the track ? A small block of mass m, which weight 7.154 N, is hit so that it slides up a frictionless inclined plane . the surface of the inclined plane is at angle theta relative to the horizontal . just after the mass is hit it has a speed of 12.80 m/s. The mass slides up the incline and travels a distance D along the incline before stating to slides before starting to slide back down . Find the work done by gravity on the black from its starting position up the ramp over the distance D. (Using g =9.80 m/s^2.) It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the breaks (without locking the wheels.) Assume that the same frictionless forces apply and find the minimum stopping distance when the car is moving at 27.0 m/s

Explanation / Answer

11)

We can solve this question using conservation of energy

Initial Kinetic energy, KEi = 0.5*mv^2

where v = 12.8 m/s

m = 7.154/9.8 =0.73 kg

Final Potential energy at the top of the ramp = mgh

where h = vertical height traveled = D

By cosnervation of energy, All KE must be converted to P.E

So, 0.5*mv^2 = mgD

So, D = 0.5*v^2/g = 0.5*12.8^2/(9.8) = 8.36 m

Work done by gravity = -mgD = -0.5*mv^2 = -0.5*(7.154/9.8)*12.8^2 = -59.8 J <-------answer

12)

Using the equation of motion,

v^2 = u^2 + 2as

where v = 0

u = 12 m/s

s = 48.96 m

So, 0^2 = 12^2 +2*a*48.96

So, a = -1.47 m/s2

NOW,

From the equatin above, s = (v^2-u^2)/(2a) = (0^2-27^2)/(2*(-1.47)) = 247.96 m <---answer

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