"on a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. wh
en max pulls on the sled with 12 n of force, directed at an angle of 15° above the ground, how much work does max do on the sled as he pulls his sister 25 m in the snow?"
1 answer:
Refer to the diagram shown below.
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
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Hi there!
We know that:
U = Potential Energy (J)
K = Kinetic Energy (J)
E = Total Energy (J)
At 10m, the total amount of energy is equivalent to:
U + K = 50 + 50 = 100 J
To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:
100J = mgh
We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².
50J = 10(10)m
m = 1/2 kg
Now, solve for height given that E = 100 J:
100J = 1/2(10)h
100J = 5h
<u>h = 20 meters</u>
Answer:
14.36 N
Explanation:
= Tension in string 1
= Tension in string 2
= mass of the bar = 2.7 kg
= weight of the bar
weight of the bar is given as
N
= mass of the bar = 1.35 kg
= weight of the monkey
weight of the monkey is given as
N
Using equilibrium of torque about left end
N
Using equilibrium of force in vertical direction
N
