The 45-kg crate is pulled up the inclined surface by the winch. The coefficient of kinetic friction between the crate and the surface is The moment of inertia of the drum on which the cable is being wound is The crate starts from rest, and the motor exerts a constant couple on the drum. Use the principle of work and energy to determine the magnitude of the velocity of the crate when it has moved 1 m.

v = 0.384 m/s

Explanation:

The complete problem is as follows:

" The 45-kg crate is pulled up the  inclined surface by the winch. The coefficient of kinetic  friction between the crate and the surface is µk = 0.4.

The moment of inertia of the drum on which the cable is  being wound is IA = 4 kg-m2

.The crate starts from rest,  and the motor exerts a constant couple M = 50 N-m  on the drum. Use the principle of work and energy to

determine the magnitude of the velocity of the crate  when it has moved 1 m "

Given:

- The mass of the crate m = 45 kg

- Coefficient of kinetic  friction µk = 0.4

- Moment of inertia of the drum IA = 4 kg-m2

- Motor constant couple M = 50 N-m

- Motor Couple moment arm R = 0.15 m

- (The missing figure is attached)

Find:

Use the principle of work and energy to

determine the magnitude of the velocity of the crate  when it has moved 1 m

Solution:

Develop a FBD of the system consisting of Motor, rope and crate. (Attachment)

   The motor generates a couple M which induces a tension in the string depending on the amount of rotation proportional to the distance moved by the block.

   The crate experiences 3 forces: Normal contact force (N) exerted by the inclined surface on the crate. The Weight (W) of the crate acting vertically downwards and Friction Force (Ff = uk*N) along the surface contact parallel to slope while opposing uphill motion. ( See Attachment )

- Apply Equilibrium condition on the crate perpendicular to the slope:

                       N - m*g*cos(20) = 0

                       N = m*g*cos(20)

- For motion of the crate as it moves 1 m along the surface the work done is given by U_12:

            U_12 = Work done by motor couple - Work done against gravity - Work done by friction

            U_12 = M*θ - m*g*sin(20) - uk*N

Where,    θ is the angle of rotation of motor through distance of 1 m = 1 / R.

            U_12 = M/R - [m*g*sin(20) - uk*m*g*cos(20)]*(1)

- The kinetic energy of system at initial condition is zero ( T_1 = 0 ). The final state Kinetic energy are as follows T_2:

            T_2 = 0.5*m*v^2 + 0.5*IA*w^2

Where,

            w = v / R  ..... Angular speed at which the motor disc rotates.(No slip)

            T_2 = 0.5*m*v^2 + 0.5*IA*v^2 / R^2

            T_2 = v^2*( 0.5*m + 0.5*IA / R^2 )

- The Total Energy balance would be:

            T_1 + U_12 = T_2

            0 + M/R - m*g*sin(20) - uk*m*g*cos(20) = v^2*( 0.5*m + 0.5*IA / R^2 )

            v^2 =[ M/R - m*g*sin(20) - uk*m*g*cos(20) ] / [ ( 0.5*m + 0.5*IA / R^2 )]

- Plug in values:

            v^2 =[ 50/0.15 - 45*9.81*sin(20) - 0.4*45*9.81*cos(20) ] / [ ( 0.5*45 + 0.5*4 / 0.15^2 )]

            v = sqrt (0.147456)

            v = 0.384 m/s

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.