You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the
system. Suppose you carry your suitcase up the stairs in 30 s while your friend takes the 60s.
Which of the following is true?
a. You did more work, but both of you expended the same power.
b. You did more work and expended more power.
c. Both of you did the same work, but you expended more power.
d. Both of you did the same work, but you expended less power.
Which of the following is true?
a. You did more work, but both of you expended the same power.
b. You did more work and expended more power.
c. Both of you did the same work, but you expended more power.
d. Both of you did the same work, but you expended less power.
1 answer:
Answer:
Both of you did the same work but you expended more power.
Explanation:
<em>Work done</em> by an object is calculated by force applied multiplied by the distance.
W=F*d
From the figure given below let us calculate force applied bith you and yopur friend.
Let us take the stairs in positive x direction,
Work done by you W₁ ,
The force applied Fₓ = F - mgsinθ =maₓ
here aₓ = 0, because both of you move with constant speed
F - mgsinθ = 0
F= mgsinθ
The work done by you on the suitcase is
W = F L cos0° , where L is he length of the staircase.
W = FL = mgsinθL , by substituting value of F
Work done by you is W₁ = mgLsinθ
Similarly work done by your friend is W₂ = mgLsinθ.
Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .
Therefore <em>work done by both of you is same</em> . Both of you did equal work.
The power , is defined as amount of energy converted or transfered per second or rate at which work is done .
P = =
Power spend by you P₁ = mgLsinθ/t
P₁ = 15*9.8*Lsinθ/30
P₁ = 4.9L sinθ eqn 1
Power spend by your friend is P₂ = mgLsinθ/t
P₂ =15*9.8*Lsinθ/60
P₂ = 2.45Lsinθ eqn 2
Dividing eqn 1 and eqn 2
P₁ = 2P₂
You have spend more power than your friend .
Hence Both of you did equal work but you spend more power.
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Answer:
A) t = 0.55 s
B) x = 24.8 m
Explanation:
A) We can find the time at which the ball will be in the air using the following equation:
Where:
is the final height= 0
is the initial height= 1.5 m
is the component of the initial speed in the vertical direction = 0 m/s
t: is the time =?
g: is the gravity = 9.81 m/s²
By solving the above equation for t we have:
Hence, the ball will stay 0.55 seconds in the air.
B) We can find the distance traveled by the ball as follows:
Where:
a: is the acceleration in the horizontal direction = 0
is the final position =?
is the initial position = 0
is the component of the initial speed in the horizontal direction = 45 m/s
Therefore, the ball will travel 24.8 meters.
I hope it helps you!