A constant horizontal force moves a 50 kg trunk 6.0 m up a 30^∘ incline at constant speed. The

step 1 Here's the free butter diagram for the trunk. We're going to apply Newton's second law in the X

A constant horizontal force moves a 50 kg trunk 6.0 m up a...

A constant horizontal force moves a $50 \mathrm{~kg}$ trunk $6.0 \mathrm{~m}$ up a $30^{\circ}$ incline at constant speed. The coefficient of kinetic friction is $0.20$. What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

Key Concepts

Constant Velocity Motion

Constant velocity implies that there is no acceleration, meaning the net force acting on the object is zero. This equilibrium condition is used to equate the applied force with the sum of opposing forces, such as the component of gravity along the incline and the kinetic friction force.

Work

Work is the energy transferred by a force acting over a distance. It is computed as the dot product of the force and displacement vectors, meaning only the component of the force in the direction of the displacement contributes to the work done.

Kinetic Friction

Kinetic friction is the resistive force that acts opposite to the direction of motion when two surfaces slide against each other. This force converts mechanical energy into thermal energy, and its magnitude is determined by the coefficient of kinetic friction multiplied by the normal force.

Inclined Plane Dynamics

Inclined plane dynamics involves analyzing forces acting on objects on a slope. This includes decomposing the gravitational force into components parallel and perpendicular to the incline, which is essential for determining both the normal force and the net force acting along the incline.

Transcript

00:01 Here's the free butter diagram for the trunk.

00:02 We're going to apply newton's second law in the x and y directions.

00:05 So we can say f sub 1 cosine of theta minus the kinetic frictional force minus mg sine of theta would be equal to the mass times the acceleration.

00:19 And then we can say in the y direction force normal minus f sub 1 sine of theta minus mg cosine of theta.

00:32 This is going to equal 0 because there is translational equilibrium in the y direction.

00:37 We know that for part a, the trunk is moving at a constant velocity, so the acceleration is going to be equal 0 meters per second squared.

00:46 And therefore, we know that the kinetic force of friction is equaling the kinetic, the coefficient of kinetic friction times the force normal.

00:56 And we can then say f sub 1 is simply going to be equal to mg sine of theta plus the coefficient of kinetic friction cosine of theta.

01:07 This would be divided by cosine of theta minus the coefficient of kinetic friction sine of theta.

01:15 And we can then say that here, the work done by the push force as the trunk is pushed through a distance l would then be equal.

01:24 We can say for part a, the work done by the pushing force sub 1 would be the force sub 1 times l cosine of theta.

01:34 This would be equal to mgl cosine of theta multiplied by sign of theta plus the coefficient of kinetic friction cosine of theta.

01:47 And this would be divided by again, cosine of theta, multiply, rather minus the coefficient of kinetic friction multiplied by sign of theta...

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