Learning Goal: To use Newton\'s laws of motion to calculate forces and accelerat

Question

Learning Goal:

To use Newton's laws of motion to calculate forces and accelerations in the motion of a particle.

In 1687, Isaac Newton presented three basic laws that describe the motion of a particle:

First law: A particle originally at rest, or moving in a straight line with a constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.

Second law: A particle acted upon by an unbalanced force, F, experiences an acceleration, a, that has the same direction as the force and a magnitude that is directly proportional to the force.

Third law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear

Newton's second law can be expressed as the formula known as the equation of motion:

F=ma

where F is the force, or sum of forces, acting on the particle; m is the mass of the particle; and a is the acceleration of the particle.

Part B

A crate sliding down a ramp reaches the bottom of the ramp and slides across a flat floor. At the instant shown, the crate has a speed of v = 6.00m/s . The crate comes to a stop after a distancex = 25.0m . (Figure 1) What is ?k, the coefficient of kinetic friction between the crate and the floor?

Express your answer numerically to three significant figures.

Part C

Two players hit an air hockey puck simultaneously. One player hits the puck with a force F1 = 5.25N at an angle ? = 25.6?. The other player hits the puck with a force F2 = 8.60N at an angle ? = 69.2?. (Figure 2) If the puck has a mass m1 = 55.0g , what is a, the acceleration of the puck, and ?, the angle of the puck's motion? The friction between the puck and the table's surface is negligible.

Express your answers numerically to three significant figures in meters per second squared and degrees separated by a comma.

Explanation / Answer

Part B)

A crate sliding down a ramp reaches the bottom of the ramp

the crate has a speed of v = 6.00m/s .

distancex = 25.0m

considering kinetic equation

v2-u2 =2aS

intial velocity of block=0m/s

force components on the block on horizontal direction

F-fk=0

ma-mu*mg=0

a=muk*g

v2 =2muk*gS

the coefficient of kinetic friction between the crate and the floor

=0.0734

Part C)

resolve force components in x-direction

y-direction

FX1= F1COS(THETA)=5.25*COS(25.6)=4.734N

FY1= F1SIN(THETA)=5.25*SIN(25.6)=2.268N

FX2= F2 COS(THETA)=8.6*COS(69.2)=3.0539N

FY2= F2SIN(THETA)=8.6*SIN(69.2)=8.0395N

resultant force along x components

assuming ball is moving in horizontal direction

so it will have horizontal accelration

7.769=ma

a=7.769/0.055

=141.25m/s2

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