Question

A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s

Answer 1

Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + ($\frac{1}{2}$ × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + ($\frac{1}{2}$ × a(2)²)

0.796  = 0 + ($\frac{1}{2}$ × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

$V_{f}$ = u + at

we substitute

$V_{f}$ = 0 + 0.398 × 2

$V_{f}$ = 0.796 m/s

now, average velocity is given as;

$V_{avg}$ = ( 0.796 m/s  + 0 ) / 2

$V_{avg}$ = ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }