A shot putter launches a 6.910 kg shot by pushing it along a straight line of length 1.650 m and at an angle of 33.80° from the horizontal, accelerating the shot to the launch speed from its initial speed of 2.500 m/s (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of 2.110 m and at an angle of 33.80°, and it lands at a horizontal distance of 15.10 m. What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)
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Vx•t = 15.1
V•cos(33.8) •t = 15.1
t=15.1/(V•cos33.8) (1)
Vy•t+ a•t²/2 = 2.11
V•sin(33.8)•t - 0.5•(9.8)•t²= -2.11 (2)
Plugging (1) into(2)
V•sin(33.2)( 15.1/(V•cos33.8) - 4.9•{15.1/(V•cos33.8) }2 = -2.11,
tan(33.8)•(15.1)- 4.9•(15.1/(V•cos33.8))2 = -2.11.
Solve for final speed V
V=11.5 m/s
Now that you have the final velocity you can now solve for the acceleration.
V ²= V₀ ²+ 2•a•x
a= (V ²- V₀ ² )/2•x =(11.5²-2.5²)/2•1.65 =38.18 m/s²
F=m•a=6.91•38.18=263.8 N
I don't think that's the right soluyion i have tried thia and i got it wrong on my assignment
Misprint Vy•t - a•t²/2 = 2.11.
The solution is correct. Check my calculations
You need to account for the force of gravity which also has to be overcome and which you ignored.
To find the magnitude of the athlete's average force on the shot during the acceleration phase, we can apply the principle of work done. The work done on an object is defined as the force applied to it multiplied by the distance over which the force is applied.
In this case, the shot putter applies a force to the shot along a straight line of length 1.650 m, at an angle of 33.80° from the horizontal. We can think of this motion as a ramp inclined at that angle. The work done by the athlete's force on the shot during the acceleration phase is given by the equation:
Work = Force * Distance * cos(θ)
Where:
- Work is the work done on the shot
- Force is the average force applied by the athlete on the shot
- Distance is the length over which the force is applied (given as 1.650 m)
- θ is the angle between the force applied and the direction of motion (given as 33.80°)
We need to find the average force, so we can rearrange the equation to solve for the force:
Force = Work / (Distance * cos(θ))
To calculate the work done on the shot, we need to know the change in kinetic energy during the acceleration phase. The initial kinetic energy of the shot is given by:
KE_initial = (1/2) * mass * (initial velocity)^2
The final kinetic energy is given by:
KE_final = (1/2) * mass * (final velocity)^2
During the acceleration phase, the shot goes from an initial velocity of 2.500 m/s to the launch velocity. The change in kinetic energy is then:
ΔKE = KE_final - KE_initial = (1/2) * mass * (launch velocity)^2 - (1/2) * mass * (2.500 m/s)^2
Now we can calculate the work done using the work-energy principle:
Work = ΔKE
Finally, substitute the values into the equation for force:
Force = Work / (Distance * cos(θ))