A block of mass m = 2.00 kg is released from rest at h = 0.300 m above the surfa

Question

A block of mass m = 2.00 kg is released from rest at h = 0.300 m above the surface of a table, at the top of a = 35.0° incline as shown in the figure 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) What time interval elapses 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

No    

Explanation / Answer

the acceleration down the incline is

a = g sin theta = 9.8 sin35 = 5.62 m/s^2

(b)

from the diagram

sin theta = h/d

d = h / sin theta = 0.3/ sin 35 = 0.52 m

the velocity of the block as it leaves the incline is

v = root 2ad = root 2 ( 5.62 m/s^2) ( 0.52 m) = 2.42 m/s

(c)

the horizontal component is

v x = v cos theta = 2.42 cos 35 = 1.98 m/s

the vertical velocity is

vy = -v sin theta= -2.42 sin 35 = -1.38m/s

(c)

the distance from the table is

R = vx t

from teh kinematic equation

H = vyt - 1/2 g t^2

-2.0 m =-1.38m/s t - 4.9 t^2

4.9 t^2+ 1.38 t - 2.0 m = 0

solving quadraatic eqution

t = 0.513 s

R= 1.98 m/s( 0.513 s) = 1.01574 m

The total time is the sum of the sliding motion and the falling motion. The time to fall is from part c and is

vf= vi + at

2.42 m/s= 0 + (5.62 m/s^2) t

t = 0.43 s

total time is T = 0.43 s +0.513 s = 0.943 s

(e)

No, the mass does not affect any result

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