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Elza [17]
3 years ago
6

Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i

t back up the 173 meter hill What amount of force did she exert on the sled?
Physics
1 answer:
Serggg [28]3 years ago
4 0

Answer:

2.2N

Explanation:

Given parameters:

Work done  = 379.5J

Height  = 173m

Unknown:

Amount of force exerted on the sled  = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

      Work done  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

     mg  = weight;

        Work done  = weight x h

           379.5 = weight  x  173

           weight  = \frac{379.5}{173}   = 2.2N

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Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

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Hence, this is the required solution.

5 0
3 years ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
3 years ago
Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done
blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=\frac{F}{m}

Hence acceleration of the block is given by:

\Rightarrow a=\frac{25}{2}=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+\frac{1}{2}at^2

Substitute the values

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Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J

3 0
3 years ago
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