A block of mass 1.58 kg which is initially at rest is acted upon by an impulse o

Question

A block of mass 1.58 kg which is initially at rest is acted upon by an impulse of 8.22 Nm due to a constant force. The impulse lasts for 0.187 s seconds and results in the mass moving at velocity v. The block then slides up a frictional ramp which has a length 2.00 m and a height 34.8 cm before colliding with a mass 3.16 kg . In this problem all horizontal surfaces are frictionless (only the incline is frictional).

If the block reaches the top of the incline with velocity 3.91 m/s , what is the coefficient of kinetic friction of the ramp?

If the 1.58 kg block collides and sticks to the 3.16 kg block, what final velocity results?

How much Kinetic Energy was lost in the collision of the two blocks?

Explanation / Answer

here,

mass , m = 1.58 kg

impulse , I = 8.22 Nm

time , t = 0.187 s

height , h = 0.348 m

length , l = 2 m

accelration by impulse , a = I/(t*m)

a = 27.82 m/s^2

v = 0 + a*t

v = 0 + 27.82 *0.187

v = 5.2 m/s

theta = arcsin(0.348/2)

theta = 10.02 degree

speed of block at the top of ramp , v' = 3.91 m/a

let the coefficient of friction be uk

using work energy theorm

uk*m*g*cos(theta)*l = 0.5 * m*v^2 - 0.5 * m*v'^2 - m*g*h

uk*9.8*cos(10.02) *2 = 0.5 * 5.2^2 - 0.5 * 3.91^2 - 9.8 * 0.348

uk = 0.128

the coefficient of kinetic friction of the rampis 0.128

let the final velocity be u

using conservation of momwnetum

(1.58 * 3.91) = ( 1.58 + 3.16) * u

u = 1.3 m/s

the final velocity is 1.3 m/s

the energy lost in the collision, KE = initial kinetic energy - final kinetic energy

KE = 0.5 * 1.58*3.91^2 - 0.5 * ( 1.58 + 3.16 ) * 1.3^2

KE = 8.07 J

the kinetic energy lost is 8.07 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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