0.400 m from the surface of a table, at the top of a -25.0° incline as shown bel

Question

0.400 m from the surface of a table, at the top of a -25.0° incline as shown below. The frictionless incline is fixed A block of mass m = 2.00 kg is released from rest at h 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? (d) What time interval elapses between when the block is released and when it hits the floor? (e) Does the mass of the block affect any of the above calculations? Yes No

Explanation / Answer

here,

mass ,m = 2 kg

h = 0.4 m

theta = 25 degree

H = 2 m

a)

the accelration of the block , a = net force /effective mass

a = ( m * g * sin(theta)) /m

a = 9.81 * sin(25) = 4.15 m/s^2

b)

the velocity of block when it leaves the incline , v = sqrt(2 * a * s)

v = sqrt(2 * 4.15 * 0.4/sin(25))

v = 2.8 m/s

c)

let the time taken to hit the floar be t

H = u * sin(theta) * t + 0.5 * g * t^2

2 = 2.8 * sin(25) * t + 0.5 * 9.81 * t^2

solving for t

t = 0.529 s

the horizontal distance from the table , x = u * cos(theta) * t

x = 2.8 * cos(25) * 0.529 m

x = 1.34 m

d)

let the time taken when traveling on inlcine be t1

v = 0 + a * t1

2.8 = 0 + 4.15 * t1

t1 = 0.67 s

the time interval elaspsed , t' = t + t1

t' = 1.2 m

e)

No,

the mass of block will not affect any of the calculations

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