a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of which hill is the kinetic energy of the roller coaster larger?

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

Related Questions

You pull your little sister across a flat snowy field on a sled. your sister plus the sled have a mass of 20 kg.the rope is at an angle of 35 degrees to the ground . You pull a distance of 50 m with a force of 30 N how much work do you do?

The workdone is is mathematically given as

W=1228.728J

Workdone

Question Parameters:

You pull your little sister across a flat snowy field on a sled. your sister plus the sled have a mass of 20 kg.the rope is at an angle of 35 degrees to the ground . You pull a distance of 50 m with a force of 30 N

Generally the equation for the horizontal force   is mathematically given as

Fcos\theta=30*cos35

Fcos\theta=24.574

distance traveled 50

Therefore

Work done is

W=24.574*50

W=1228.728J

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Explanation:

Given

Combined mass(m)= 20 kg

rope is at an angle of

Distance pulled =50 N

Force= 30 N

there will be two components of force

and

will do work while the work will be as angle between force and displacement is zero

A baseball pitcher throws a ball with a speed of 41 m/s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates the ball through a displacement of about 3.5 m, from behind the body to the point where it is released.

When t=0, v=0, d=0 When t=tf, v=41m/s, d=3.5m We have 2 formulas – the ones corresponding to uniformly accelerated linear movement: vf=a*t+vo d=(1/2)*a*t^2+vo*t Let’s put the data in the formulas: 41m/s=a*t+0=a*t 3.5m=(1/2)*a*t^2+0*t=1/2*a*t^2 You can use a variety of methods to find t and a. I will choose substitution. t=(41m/s)/a 3.5m=(1/2)*a*((41m/s)/a)^2=(1/2)*a*(41m/s)^2/a^2=(1/2)*(41m/s)^2/a a=(1/2)*(41m/s)^2/(3.5m)=(1/2)*41^2(m^2/s^2)/(3.5m) a=41^2(m/s^2)/( 2*3.5)=240m/s^2

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