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3 years ago

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Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

om 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

2 answers:

slava [35]3 years ago

7 0

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

photoshop1234 [79]3 years ago

7 0

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

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Two 2.3 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . What are (a) the x-com

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Missing details in the question:

The velocities before collision are

$u_A=(40i+49j) m/s$

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After the collision:

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(a) 61 m/s

We can solve the problem by simply treating separately the x- and the y-components of the motion.

Here we want to analzye the motion along x. We have:

$u_A = 40 m/s$ is the initial velocity of A along the x-direction

$u_B = 35 m/s$ is the initial velocity of B along the x-direction

$v_A = 14 m/s$ is the final velocity of A along the x-direction

$v_B = ?$ is the final velocity of B along the x-direction

Since the total momentum along the x-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

where

m = 2.3 kg is the mass of the two bodies. Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the x-direction:

$v_B = u_A + u_B - v_A=40+35-14=61 m/s$

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Similarly to what we did in part a), here we analyze the conservation of momentum along the y-direction.

We have:

$u_A = 49 m/s$ is the initial velocity of A along the y-direction

$u_B = 11 m/s$ is the initial velocity of B along the y-direction

$v_A = 21 m/s$ is the final velocity of A along the y-direction

$v_B = ?$ is the final velocity of B along the y-direction

Since the total momentum along the y-direction must be conserved, we can write

$mu_A + mu_B = mv_A + mv_B$

Since the mass is the same, we can eliminate it from the equation,

$u_A + u_B = v_A + v_B$

And so, we find the final velocity of B along the y-direction:

$v_B = u_A + u_B - v_A=49+11-21=39 m/s$

c) +615 J

Here we have to find the total kinetic energy before and after the collision first.

First, we have to find the speed of each object before and after the collision. We have:

$u_A = \sqrt{40^2+49^2}=63.2 m/s\\u_B = \sqrt{35^2+11^2}=36.7 m/s\\v_A = \sqrt{14^2+21^2}=25.2 m/s\\v_B = \sqrt{61^2+39^2}=72.4 m/s$

So, the total kinetic energy before the collision was

$K_i = \frac{1}{2}mu_A^2+\frac{1}{2}mu_B^2 = \frac{1}{2}(2.3)(63.2)^2+\frac{1}{2}(2.3)(36.7)^2=6143 J$

While after the collision

$K_f = \frac{1}{2}mv_A^2+\frac{1}{2}mv_B^2 = \frac{1}{2}(2.3)(25.2)^2+\frac{1}{2}(2.3)(72.4)^2=6758 J$

So, the change in kinetic energy is

$\Delta K = K_f - K_i = 6758-6143 = +615 J$

(note that the system cannot gain kinetic energy in the collision, unless there is an external force acting on it)

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