Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
1 answer:
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
where
is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
where
is the final speed of the sled
is the initial speed
is the distance covered
By rearranging the equation, we find d:
where
Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
where
By rearranging the equation, we find d:
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Answer:
80%
Explanation:
Efficiency = Power output / Power input × 100 %
To calculate efficiency we need to find power output of electric pump.
We can use,
Work done = Energy change
Work done per second = Energy change per second
Work done per second = Power
Therefore, Power = Energy change per second
= Change in potential energy of water per second
=mgh / t
= 200× 10×6 / 10
= 1200 W = 1.2 kW
Now use the first equation to find efficiency,
Efficiency = × 100%
= 80 %
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s