Question

A 2-kg bowling ball sits on top of a building that is 40 meters tall.

Answer 1

1. A 2-kg bowling ball sits on top of a building that is 40 meters tall. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has:
KE = 0 (because velocity is 0)
PE = mgh = 2kg*9.8m/s^2*40m= 784 joule
2. A 2-kg bowling ball rolls at a speed of 5 m/s on the roof of the building that is 40 tall. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has:
BothKE = [1/2]m*v^2 = [1/2](2kg)(5m/s)^2 =  25 joulPE = mgh = 40 joule
3. A 2-kg bowling ball rolls at a speed of 10 m/s on the ground. Circle one: KE / GPE / both Show your work for finding the values of each type of energy the object has: 
KE = [1/2]m(v^2) = [1/2]2kg*(10m/s)^2 = 100 joule
GPE = mgh = 0
1. A 1,000-kg car has 50,000 joules of kinetic energy. What is its speed?
KE = [1/2]m(v^2) => v = √[2KE/m] = √[2*50,000joules/1000kg] = 10m/s
2. A 200-kg boulder has 39,200 joules of gravitational potential energy. What height is it at? 
GPE = mgh => h = GPE / (mg) = 39,200 joules / (200kg * 9.8m/s^2) = 20m
3. A 1-kg model airplane has 12.5 joules of kinetic energy and 98 joules of gravitational potential energy. What is its speed? What is its height?
KE = [1/2]m(v^2) => v = √ [ 2KE/m] = √[2*12.5 j / 1kg] = 7.9 m/s
GPE = mgh => h = GPE/(mg) = 98/(1kg*9.8m/s^2) = 10 m

Answer 2

Answer:

The GPE:

So, each object, matter or system holds some stored energy inside its molecules and such energy is stored in the system due to its position or height from the surface or ground.

Mathematical equation:

GPE=mgh, (while "m" is the mass, "g" is gravity, and "h" is the height of the object where it is been placed on).

The K.E(Kinetic energy):

Kinetic energy is the amount energy produced or possessed by it due to some motion in any direction.

Mathematical equation:

K.E= 1/2 m×v², (As "m" is the mass of the body or object, while the velocity represents its motion)

Explanation:

1.Answer-(G.P.E):Because, it is not in motion but it is been placed on some point at a certain height,h from the ground or surface.

So, we have, G.P.E= m.g.h and by putting the values into the equation, we can get the results as follow,

G.P.E=(2)×(9.8)×(40)⇒ G.P.E=  784 kg.m².sec²,

2.Answer-(Both): The body is present on the roof top at height  giving it the G.P.E possessed by the object or the body, along with the K.E due to its moment on the top of the building.

Calculation: K.E+G.P.E= Total energy⇒ Tot. energy= 1/2×(2)×(5)²+ (2)×(9.8)×(40),

Tot.energy= 809 J(kg.m².sec²).

3.Answer-(K.E):

When the ball is present on the ground and it is in motion(as it h=0). Then the energy possessed by the body will be Kinetic energy,K.E.

Calculations: K.E=1/2×(2)×(10)²,

K.E=1/2×(2)×(100),

K.E=100 kg.m².sec².

3.Answer-(Both, but at the summit point it will be G.P.E and during the downfall it will be K.E): When a child is thrown upwards in the air by his or her parents then the G.P.E will be attained by the subject at a certain point called as the summit point, and then during the down fall it will change into the K.E.

Then, Tot. energy= m.g.h + 1/2 m.v²,

⇒ Tot.energy= (20)×(9.8)×(2) + 1/2 (20)×(5)²,

⇒Tot.energy=392+250,

⇒Tot.energy=642 kg.m².sec².

• Part:2,

As, we know that the total G.P.E possessed by any object is equal to the total K.E, then we can say that;

• K.E=P.E,

But,along with that we can write as,

• K.E=1/2×(1000)×(v)²,

Putting the required values in the above equation we can have,

• 50,000=1/2×(1000)×(v)²,
• 50,000=500×(v)²,
• 50,000/500=v²,
• v²=100,
• √v²=√100,
• v=10 m/sec,

Now, to find the height,h=?

• G.P.E=m.g.h,
• 39,200=(200)×(9.8)×(h),
• 39,200=1960×(h),
• h=20 meter,

As to find the speed,v=? and the height,h=?.

We, have for speed the following formula:

K.E=1/2×(m)×(v)²,

12.5=1/2×(1)×(v)²,

v²=25,

so,

√v²=√25,

v=5 m/sec,

And to find the height,h=?, we have;

G.P.E=m.g.h,

98=(1)×(9.8)×(h),

h=10 meters.