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AURORKA [14]
3 years ago
11

An average person can reach a maximum height of about 60 {\rm cm} when jumping straight upfrom a crouched position. During the j

ump itself, the person's bodyfrom the knees up typically rises a distance of around 50 {\rm cm}. To keep the calculationssimple and yet get a reasonable result, assume that the entirebody rises this much during the jump. In terms of this jumper's weight W , what force does the ground exert on him or herduring the jump?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

2.2 W where W is the weight of the jumper.

Explanation:

A) use the formula

V^2 = 2gh= 2*9.81*60

that gives

V = 3.43 m/s

b) W = mg

during the jump , the displacement of the body is d =0.5m

so the acceleration a = (V^2-Vo^2)/2d = 11.76 m/s^2

net force applied by ground = m(g+a) = W/g(g+a)

= 2.2 W

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An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 5 N, F2 = 8 N, and F3 = 5
DIA [1.3K]

The figure is missing: find it in attachment.

a) (3i - 5j) N

First of all, we have to write each force as a vector with a direction:

- Force F1 points downward, so along the negative y-direction, so we can write it as

F_1 = -5 j

- Force F2 points to the right, so along the positive x-direction, so we can write it as

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- Force F3 points to the left, so along the negative x-direction, so we can write it as

F_3 = -5 i

Now we can write the net force, by adding the three vectors and separating the x-component from the y-component:

F=F_1+F_2+F_3 = -5j+8i-5i = (8-5)i+(-5)j=(3i-5j)N

b) 5.8 N

The magnitude of the net force can be calculated by applying Pythagorean's theorem on the components of the net force:

F=\sqrt{F_x^2+F_y^2}

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation,

F=\sqrt{(3)^2+(-5)^2}=5.8 N

c) -59.0^{\circ}

The angle of the net force, measured with respect to the positive x-axis (counterclockwise), can be calculated by using the formula

\theta=tan^{-1} (\frac{F_y}{F_x})

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation, we find

\theta = tan^{-1} (\frac{-5}{3})=-59.0^{\circ}

d) 0.84 m/s^2

The acceleration can be found by using Newton's second law:

F=ma

where

F is the net force on an object

m is its mass

a is the acceleration

For the object in the problem, we know

F = 5.8 N

m = 6.9 kg

Solving the equation for a, we find the magnitude of the acceleration:

a=\frac{F}{m}=\frac{5.8}{6.9}=0.84 m/s^2

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