. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

Substituting 42° for θ and 87.0° for

Meanwhile, the vertical component is:

Again substituting 42° for θ and 87.0° for

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

Where;

is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting . Normal reaction becomes:

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

Rearranging the equation above to isolate leads to:

Substituting in the following values:

m = 73 kg

g = 9.8 m/s²

Thus:

Therefore, the coefficient of friction is = 0.0984

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change

initial speed

Final Speed

Specific internal Energy

according to the first Law of thermodynamics

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.

The resulting motion can be determined using the Pythagorean theorem, as the two components (north and east) are at right angles. To find the direction, trigonometry is applied, yielding Ф=arctan(3.8/12)=17.57° north of east.