I need help with 2 physics questions, I just want to know what equation to use to solve the problems.A 55kg person is skating. She reaches a maximum speed and then glides to a stop in 100 m. The...

I need help with 2 physics questions, I just want to know what equation to use to solve the problems.

A 55kg person is skating. She reaches a maximum speed and then glides to a stop in 100 m. The coefficient between her skates and the ice is 0.010. What is the work done by the kinetic friction?

A roller coaster has 2 vertical loops one after another. The roller coaster has a speed of 7.00 m/s at the top of the first loop with a height of 22.2m. It then proceeds around the second vertical loop with a height of 15.0m. What is the speed of the roller coaster at the top of the second vertical loop?

Asked on by jessica35

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maadhav19's profile pic

maadhav19 | College Teacher | (Level 2) Assistant Educator

Posted on

For the first problem, recall that work is force times distance:   W=F*r

In this case, the force is the force of friction, and can be calculated as the coefficient of friction times the mass times acceleration due to gravity: F=u*m*g

 

For the second problem, the total energy, potential and kinetic, is conserved. So the potential plus kinetic energy at the top of the first loop equals the potential plus kinetic energy at the top of the second one:

K1+U1 = K2+U2

Where kinetic energy K = (1/2)mv^2 and potential energy U=mgh.

m= mass (notice what happens to it when you write out these eqs)

v^2 is the velocity squared

g is the acceleration due to gravity

h is the height of each loop.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

a) Force,F due to frction is given by:

F= M*(mg) , where M (or mu) is the coefficient of kinetic friction .

The distance through which force of kinetic friction worked = 100m.

Therefore, the work done by the kinetic friction worked =100 m

So the work done by the kinetic frictional force = M(mg)*100 = 0.1(55kg*g meter/second ^2)*100m =0.10*55*9.81*100 =5395.5meter^2/second^2

=5395.5 (meter/secon^2 )meter

=5395.5 Newton meter

=5395.5 Joules of work done by the kinetic friction.

2)

The roller coasetr has two types of energy at any height- (1) Potential energy and (ii) Kinetic energy. The potential energy is determined by the height, h of the roller coaster and the kinetic energy is determined by the velocity, v of the roller coaster at any moment. But the sum of the PE and KE is always a constant according to the conservation principle of the energy.

So, mgh+(1/2)mv^2 = k, a constant, where m is mass of the roller coaster and g is the acceleration dueto gravity. Putting values, we get:

m kg*9.81 meter/sec^2 *22.2meter +(1/2)m*(7meter)^2= k = m kg*g*15meter* (1/2)m(v' meter)^2, where v' is the velocity of the roller coaster at the height 15 meter and is to be found out. So, solving for v' , we get the required velocity.

So, cancelling the commons,

[9.81*22.2+(1/2)*7^2] (meter/sec)^2= 9.81*15 (meter/sec)^2+(1/2)v'^2.

(1/2)v'^2 = [p9.81(22.2-15)+(1/2)*7^2 ] (m/s) ^2or

v'^2 = 2*9.81*7.2 + 7^2 =190.264 (m/s)^2 or

v' = 13.79.36 m/s

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