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 = 25.0° 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?YesNo

Explanation / Answer

here,

mass , m = 2 kg

h = 0.5 m

theta = 25 degree

H = 2 m

(a)

the accelratiion of the block , a = g * sin(theta)

a = 9.8 * sin(25)

a = 4.14 m/s^2

the accelration of the block is 4.14 m/s^2

(b)

let the velocity of block as it leaves the incline be v0

using conservation of energy

0.5 * m * v0^2 = m * g * h

0.5 * v0^2 = 9.8 * 0.5

v0 = 3.13 m/s

the velocity of block as it leaves the incline is 3.13 m/s

(c)

let the time taken to hit the floar from the table be t1

H = v*sin(theta) * t1 + 0.5 * g * t1^2

2 = 3.13 * 0.4226 * t1 + 0.5 * 9.8 * t1^2

t1 = 0.52 s

the distnace of block from the table when it hits the floar , R = v*cos(theta) * t1

R = 3.13 * cos(25) * 0.52

R = 1.48 m

the distnace of block from the table when it hits the floar is 1.48 m

(d)

let the time for block on incline be t2

using first equation of motion

v = u + a*t2

3.13 = 0 + 9.8 * sin(25) * t2

t2 = 0.76 s

the total time elapsed is t1 + t2

the time elapsed between when the block is released and when it hits the floor is 1.28 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.