A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surfac

Question

A block of mass m = 2.00 kg is released from rest at h = 0.800 m from the surface of a table, at the top of a = 35.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

(a) Determine the acceleration of the block as it slides down the incline.
_____m/s2

(b) What is the velocity of the block as it leaves the incline?
_____m/s

(c) How far from the table will the block hit the floor?
_____m

(d) How much time has elapsed between when the block is released and when it hits the floor?
____s
(e) Does the mass of the block affect any of the above calculations?

Yes or No    

Explanation / Answer

A. the acceleration down the incline is given as

a = g*sinA = 9.8*sin 35 deg = 5.62 m/sec^2

B. v = (2*a*d)^0.5 = (2*5.62*(1.394))^0.5 = 3.95 m/sec

where, sin A = h/d so d = h/sinA .

C. V = 3.95 m/sec

Vx = 3.95*cos 35 deg = 3.235 m/sec

Vy = -3.95*sin 35 deg = -2.265 m/sec

R = Vx*t = 3.235*t

H = -2 = -2.265*t - 4.9*t^2

t = 0.448 sec

R = 3.235*0.448 = 1.449 m

D. The total time is the sum of the sliding motion and the falling motion. The time to fall is from part c is 0.448 sec. The time to slide is found from

vf = vi + at

t2 = 3.95/4.9 = 0.806 sec.

Therefore the total time from release to landing is

t = t1 + t2

t = 0.448 + 0.806 = 1.254 sec

E. No, the mass does not affect any result since it does not appear in any of the above calculations

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