Question

A 400 N student climbs up a 3.0 m ladder in 4.0s.

Answer 1

Answer:

We are given that the student has a weight of 400N

So, the force applied by the boy has to be added to 400 since he has to apply force to keep his body on the stairs

We are given:

u = 0 m/s

v = v m/s

t = 4 s

a = a m/s/s

s = 3 m

w = 400 N

Converting to Mass

w = m * g    (where m is the actual mass and g is the value of gravity)

letting g = 10m/s/s

400  = 10 * m

m = 40 kg

Finding Acceleration:

From the second equation of motion:

s = ut + (a*t*t) / 2

3 = (0)(4) + (a)(4)(4)/2

3 = 8a

a = 3/8 m/s/s

Force Applied by the Student:

Now that we have the acceleration, we can find the force applied by the student to go up the stairs

F = ma (Newton's second law of motion)

F = (40)(3/8)

F = 15 N

Hence, the student is applying 15N to climb up the stairs and another 400 N to support his mass

Total Force applied by the Student:

F(total) = F(climbing) + F(weight)

F(total) = 15 + 400

F(total) = 415 N

Work done by the Student:

Work done = F.S   (dot product of force applied and displacement)

Work done = (415)(3)

Work done = 1245 Joules

Power Output of the Student:

power output is the amount of work done per second

Power = Work Done / t

Power = 1245 / 4

Power = 311.25 Watt

**note that the this is just the power consumed by the student to climb the stairs**