A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surfac

Question

A block of mass m = 2.00 kg is released from rest at h = 0.500 m from the surface of a table, at the top of a = 20.0° 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 No    

Explanation / Answer

Here ,

mass , m = 2 Kg

a)

as the incline is friction less

acceleration of block down the incline , ai = g * sin(theta)

ai = 9.8 * sin(20)

ai = 3.351 m/s^2

b)

let the velocity is uf

using third equation of motion

v^2 - u^2 = 2 * a * d

uf^2 - 0 = 2 * 3.351 * (0.5/sin(20))

uf = 3.13 m/s

the velocity of block when it leaves the incline is 3.13 s

c)

let the time of fall is t

Using seocnd equation of motion in vertical direction

y = u * t + 0.5 at^2

2 = 3.13 * sin(20) * t + 0.5 * 9.8 * t^2

t = 0.54 s

distance from table = 0.54 * 3.13 * cos(20)

distance from table = 1.59 m

the block will hit 1.59 m away from table

d)

time taken is 0.54 s

e)

NO , the mass of block does not affect any of these

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