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Sholpan [36]
4 years ago
8

A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

done on the object?
Physics
2 answers:
gtnhenbr [62]4 years ago
8 0

<u>Answer:</u> The work done on the object is 110 W

<u>Explanation:</u>

Work is defined as the force that is causing the movement or displacement of an object.

Mathematically,

W=F\times d

where,

W = work done on the object

F = Force applied on the object = 55 N

d = displacement = 10 m

Putting values in above equation, we get:

W=55N\times 2m\\\\W=110N.m=110W

Hence, the work done on the object is 110 W

Elanso [62]4 years ago
6 0
F=55 \ N \\ d=10 \ m \\ \boxed{W-?} \\ \bold{Solving:} \\ \boxed{W=F \cdot d} \\ W=55\ N \cdot 10 \ m \\ \Rightarrow \boxed{W=550 \ J}
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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. It gets blocked just after re
brilliants [131]
1) Data:

Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m

2) Questions: V at Y = 3m and X at Y = 3 m

3) Calculate components of the initial velocity

Vox = Vo * cos(37°) = 15.97 m/s

Voy = Vo * sin(37°) = 12.04 m/s

4) Formulas

Vx = constant = 15.97 m/s

X = Vx * t

Vy = Voy - g*t

Y = Yo + Voy * t - g (t^2) / 2

5) Calculate t when Y = 3m (first time)

Use g ≈ 9.8 m/s^2

3 = 12.04 * t - 4.9 t^2

=> 4.9 t^2 - 12.04t + 3 = 0

Use the quadratic equation to solve the equation

=> t = 0.28 s and t = 2.18s

First time => t = 0.28 s.

6) Calculate Vy when t = 0.28 s

Vy = 12.04 m/s  - 9.8 * 0.28s = 9.3 m/s

7) Calculate V:

V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s

tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°

Answer: V ≈ 18.5 m/s, with angle ≈ 30°

8) Calculate X at t = 0.28s

X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m

Answer: X ≈ 4,5 m
4 0
3 years ago
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