An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
1 answer:
a) Average power: 1425 W
b) Instantaneous power at 3.0 sec: 2850 W
Explanation:
a)
The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation
where
s = 18 m is the displacement along the ramp
u = 0 is the initial velocity
t = 3.0 s is the time taken
a is the acceleration of the object along the ramp
Solving for a,
Now we can apply Newton's second law to find the net force on the object:
This net force is the resultant of the applied force forward () and the component of the weight acting backward (
), so we can find what is the applied force:
where
m = 24 kg is the mass of the object
is the acceleration of gravity
Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:
where s = 18 m is the displacement.
Therefore the average power needed is:
b)
The instantaneous power at any point of the motion is given by
where
is the force applied
v is the velocity of the object
We already calculated the applied force:
While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:
And therefore, the instantaeous power at 3.0 sec is:
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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Complete Question
The complete question is shown on the first uploaded image
Answer:
Force on each hand is 196.22 N
Force on each foot is 95.8 N
Explanation:
In order to get a better understanding of this question let us explain some concepts
Normal Force:
We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.
Torque:
We can define torque as the moment of forces that tends to produce or cause rotation
From the question we are given that
Weight of body is (W) = 584 N
The normal force on both hands (Ha) = ?
The normal force on both legs (Lg) = ?
Looking at the diagram the person is at equilibrium so
584 = Ha + Lg
an also this mean that torques acting on the body is balanced
So, 0.410 Ha = 0.840 Lg
Making Lg the subject of formula in the equation above we
Lg = 0.4881 Ha
Considering the first equation and replacing Lg with this recent equation we have
584 = Ha + 0.4881 Ha
Therefore Ha = 392.44 N
This value obtained is for both hands for each hand we divide by 2
Therefore we have for each hand =196.55 N
Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have
Lg = 0.4881 × 392.44
= 191.22 N
The value above is the force on both legs to obtain the force on each leg we have
191.22/2 = 95.8 N.
