Question

What if? if the ramp is 4. 00 m long, what is the maximum coefficient of friction that would allow the block to reach the end of the ramp?.

Answer 1

The maximum coefficient of friction that would allow the block to reach the end of the ramp is equal to 0.3611.

Given the following data:

• Mass = 195 grams to kg = 0.195 kilogram.

• Force constant = 1.60 kN/m to N/m = 1,600 N/m.

• Extension = 10.0 cm to m = 0.1 meter.

• Length of ramp = 4.00 meters.

• Angle of inclination = 60.0°.

Scientific data:

• Acceleration due to gravity on Earth = 9.8 $m/s^2$

How to calculate the maximum coefficient of friction.

Since the ramp obeys Hooke’s law and the length (distance) of the ramp is known, we would apply this formula:

$\frac{1}{2} kx^2=mgdsin \theta +umgdcos\theta$

Where:

• k is the force constant.

• m is the mass.

• g is acceleration due to gravity.

• x is the extension.

• u is the coefficient of friction.

Making u the subject of formula, we have:

$\frac{1}{2} kx^2=mgdsin \theta +umgdcos\theta\\\\umgdcos\theta = \frac{1}{2} kx^2 - mgdsin \theta\\\\2umgdcos\theta = kx^2 - 2mgdsin \theta\\\\u=\frac{kx^2}{2mgdcos\theta} -\frac{2mgdsin \theta}{2mgdcos\theta} \\\\u=\frac{kx^2}{2mgdcos\theta} -Tan\theta$

Substituting the given parameters into the formula, we have;

$u=\frac{1600 \times 0.1^2}{2 \times 0.195 \times 9.8 \times 4.00 \times cos60} - Tan60\\\\u=\frac{1600 \times 0.01}{2 \times 0.195 \times 9.8 \times 4.00 \times 0.5} - 1.7321\\\\u=\frac{16}{7.644} - 1.7321\\\\u=2.0932 - 1.7321$

u = 0.3611.

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Complete Question:

A 195 g block is pressed against a spring of force constant 1.55 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. What if? if the ramp is 4. 00 m long, what is the maximum coefficient of friction that would allow the block to reach the end of the ramp?