crosez
contestada

A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and friction.

A) After the cyclist has traveled 180 m , what was the net work done by gravity on the cyclist?

B) How fast is the cyclist going? in m/s

please explain if possible!!

Respuesta :

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

abidemiokin

The network done by gravity on the cyclist is 10,459.28 Joules

The velocity of the cyclist in m/s is 15.69m/s

The formula for calculating the workdone is expressed using the formula:

[tex]W=Fdsin \theta[/tex]

Since F = mg

[tex]W =mgsin \theta[/tex]

m is the mass of the bicycle

g is the acceleration due to gravity

[tex]\theta[/tex] is the angle of inclination

Given the following parameter

m = 85 kg

g = 9.8m/s²

[tex]\theta[/tex] = 4.0 degrees

A) Substitute the given data into the formula as shown;

[tex]W =85(18)\times 180sin 4^0\\W=833\times 180sin4^0\\W=10,459.28 Joules[/tex]

Hence the net work done by gravity on the cyclist is 10,459.28 Joules

B) The kinetic energy is the energy possessed by a body by virtue of its motion.

The formula for calculating the KE is expressed as;

[tex]KE =\frac{1}{2}mv^2\\[/tex]

[tex]10,459.28 = \frac{1}{2}\times 85 \times v^2\\85v^2= 20,918.57\\v^2=\frac{20,918.57}{85}\\v^2= 246.10\\v=\sqrt{246.10}\\v= 15.69m/s[/tex]

Hence the velocity of the cyclist in m/s is 15.69m/s

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