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    

ANSWERS:

a)5.62m/s^2

b)3.97m/s

c)1.43 m

d) ??????

/no

PLease i need help on letter d

Explanation / Answer

Here ,

theta = 35 degree

h = 0.8 m

H = 2 m

a) for the acceleration , a = g * sin(theta)

a = 9.8 * sin(35)

a = 5.62 m/s^2

the acceleration of the blocks is 5.62 m/s^2

b)

velocity is v

v = sqrt(2 * g * h)

v = sqrt(2 * 9.8 * 0.8)

v = 1.42 m/s

c)

let the time of flight is tf

2 = 1.42 * sin(35) * t + 0.5 * 9.8 * t^2

solving for t

t = 0.56 s

distance = 0.56 * 1.42 * cos(35)

distance = 0.652 m

the block will hit 0.652 m away

d)

time taken is 0.652 m

e)

as there is no mass in any of calculations ,

the mass of block does not affect

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