In a 70-degree F classroom, a 4-kg aluminum block begins its adventure at the to

Question

In a 70-degree F classroom, a 4-kg aluminum block begins its adventure at the top of a 1-m radius ramp that is rough enough to provide a constant 2 N frictional force. At the bottom of the ramp, it collides and latches with a 1 kg copper block. The pair then slide along 1 m further on the rough surface before lauching off the edge of the ramp, falling to the floor 3.2 m below.

A.) Determine the speed of the 4-kg block justbefore it collides with the 1-kg block.

B.) If all the energy dissipated by friction just before it collides with the 1-kg block goes to heat the aluminum block, determine the change in temperature in the aluminum (Specific heat of aluminum 900 J/kg).

C.) Determine the speed of the pair of blocks after the collision.

D.) Determine the speed of the pair of blocks at launch.

E.) How far from the edge of the ramp do the blocks land?.

Explanation / Answer

A) let V is the speed at the bottom.

Workdone on aluminium bock, W = m*g*R - 2*pi*R/4

= 4*9.8*1 - 2*3.14*1/4

= 37.63 J

According to Work-Enrgy therom,

W = 0.5*m*(v2^2-u^2)

here, u = 0

W = 0.5*m*v^2

v = sqrt(2*W/m)

= sqrt(2*37.63/4)

= 4.34 m/s <<<<<<<<<<---------Answer

B) Q = m*g*h

= 4*9.8*1

= 39.2 J

Q = m*C*dT

dT = Q/(m*C)

= 39.2/(4*900)

= 0.01 degrees celsius <<<<<<<<<<---------Answer

C)

m1*v1 = (m1+m2)*V

V = m1*v1/(m1+m2)

= 4*4.34/(4+1)

= 3.47 m/s <<<<<<<<<<---------Answer

D)a = -F/(m1+m2)

= -2/(4+1)

= -0.4 m/s^2

let V' is the spped before launching

v'^2 - V^2 = 2*a*s

V'^2 = V^2 +2*a*s

V'^2 = 3.47^2 -2*0.4*1

V' = 3.35 m/s <<<<<<<<<<---------Answer

E) let t is the time taken,

H = 0.5*g*t^2

t = sqrt(2*H/g)

= sqrt(2*3.2/9.8)

= 0.808 m

X = V'*t

= 3.35*0.808

= 2.7 m <<<<<<<<<<---------Answer

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