A factory worker accidentally releases a 180 kg crate that was being held at rest at the top of

step 1 Well, mass of crate M is equal to 180 kilograms. Mass M is equal to 180 kilograms, 180 kilograms

A factory worker accidentally releases a 180 kg crate that...

A factory worker accidentally releases a $180 \mathrm{~kg}$ crate that was being held at rest at the top of a ramp that is $3.7 \mathrm{~m}$ long and inclined at $39^{\circ}$ to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is $0.28$. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

Key Concepts

Conservation of Energy

This concept involves tracking how energy transforms from one form to another within a system. In problems like this, gravitational potential energy is converted into kinetic energy as the object moves, with some energy lost to friction. The principle helps in setting up equations that relate the change in potential energy to the gain in kinetic energy and the work done by non-conservative forces.

Work Done by Friction

Friction is a non-conservative force that dissipates energy as heat. In this context, work done by friction is calculated by multiplying the friction force and the distance over which it acts, and it is subtracted from the mechanical energy available for conversion to kinetic energy. This calculation is essential to determine the object’s speed and its motion on different surfaces.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its height above a reference level. It is calculated using the mass, gravitational acceleration, and the vertical displacement. This energy is the source used to overcome friction and provide kinetic energy as the object moves down the incline.

Kinetic Friction

Kinetic friction is the resisting force acting between surfaces in relative motion and is proportional to the normal force and the coefficient of kinetic friction. Its role is crucial in determining how much energy is lost during motion, affecting the final speed and distance traveled by the object.

Force Decomposition on an Incline

When dealing with inclined planes, gravitational force is broken down into components parallel to and perpendicular to the incline. The parallel component controls the acceleration down the slope while the perpendicular component determines the normal force, which in turn affects the friction force. This decomposition is essential for accurate energy and motion analysis.

Mass Independence in Energy Analysis

In systems where friction is proportional to the normal force, the mass of the object often cancels out in the energy equations. This leads to outcomes such as the final speed and sliding distance being independent of mass, given that other factors like the coefficient of friction and ramp geometry remain constant.

Transcript

00:01 Well, mass of crate m is equal to 180 kilograms.

00:04 Mass m is equal to 180 kilograms, 180 kilograms.

00:12 And theta is equal to 39 degrees.

00:18 And length of the ram, l is equal to 3 .7 meters.

00:28 And the coefficient of kinetic friction, mu k, is equal to 0 .28.

00:38 Okay, now let's solve part a.

00:43 According to conservation of energy, kinetic energy at top plus potential energy at top is equal to kinetic energy at the bottom plus potential energy at the bottom plus the thermal energy or the thermal energy of the kinetic friction at the bottom bottom and this is equation number one we have normal force are and it's equal to m g cosine theta and it's equal to 180 multiply by 9 .8 multiply by cosine of 39 degrees and it's equal to 1370 .1370 .89 newton's.

01:50 0 .89 newton's is the normal force.

01:57 And the vertical distance h equals h is equal to l sine theta sine theta l is 3 .7 multiply by sine of 39 degrees and it's equal to 2 .33 meters and now substituting the values in equation number one we have 1 divided by 2 times m 0 square plus m g h is equal to half m vb square plus 0 plus mu k times r times l and it's equal to 180 multiply by 9 .8 multiply by 2 .33, 2 .33.

03:15 And solving this further, we have 1 divided by 2, multiply by 180, 180, multiply by wb square, plus 0 .282, 0 .28 ,000, multiply by wb square, plus 0 .28 0 .28 multiplied by 1370 .89 multiply by 3 .7.

03:54 3 .7 and it's equal to 4 -110 .1 .12...

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