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edexcel m1 - newton's f = ma law question help..
Albert and Bella are both standing in a lift. The mass of the lift is 250g. 1. As the lift moves upward with a constant acceleration, the floor of the lift exerts forces of magnitude 678N and 452N respectively on Albert and Bella. The tension in the cable which is pulling the lift upwards is 3955N.
(a) Find the acceleration of the lift
ite.. so the total upward force is 3955 (tension on cable) + 678 + 452 (reactions on bert and bella) = 5085. the total downward force is 250 x 9.8 = 2450.
Resultant force = Mass x Acceleration
5085 - 2450 = 250a
a = 2635/250 = 10.54ms^-2
However, the book gives me the answer 1.5ms^-2, where could i have gone wrong?
2.The engine of a van of mass 400kg cuts out when it is moving along a straight horizontal road with speed 16ms^-1. The van comes to rest without the breaks being applied.
In a model of the situation, it is assumed that the van is subject to a resistive force which has constant magnitude of 200N.
(a) Find how long it takes the van to stop
ite. so:
Force = Mass x Acceleration
Acceleration = (v-u)/t = (0-16)/t
Mass = 400
Force = R - 200
but now we have 2 unknowns, R and t. what has to be done here?
(a) Find the acceleration of the lift
ite.. so the total upward force is 3955 (tension on cable) + 678 + 452 (reactions on bert and bella) = 5085. the total downward force is 250 x 9.8 = 2450.
Resultant force = Mass x Acceleration
5085 - 2450 = 250a
a = 2635/250 = 10.54ms^-2
However, the book gives me the answer 1.5ms^-2, where could i have gone wrong?
2.The engine of a van of mass 400kg cuts out when it is moving along a straight horizontal road with speed 16ms^-1. The van comes to rest without the breaks being applied.
In a model of the situation, it is assumed that the van is subject to a resistive force which has constant magnitude of 200N.
(a) Find how long it takes the van to stop
ite. so:
Force = Mass x Acceleration
Acceleration = (v-u)/t = (0-16)/t
Mass = 400
Force = R - 200
but now we have 2 unknowns, R and t. what has to be done here?
Scroll to see replies
dumb maths student
2.The engine of a van of mass 400kg cuts out when it is moving along a straight horizontal road with speed 16ms^-1. The van comes to rest without the breaks being applied.
In a model of the situation, it is assumed that the van is subject to a resistive force which has constant magnitude of 200N.
(a) Find how long it takes the van to stop
ite. so:
Force = Mass x Acceleration
Acceleration = (v-u)/t = (0-16)/t
Mass = 400
Force = R - 200
but now we have 2 unknowns, R and t. what has to be done here?
2.The engine of a van of mass 400kg cuts out when it is moving along a straight horizontal road with speed 16ms^-1. The van comes to rest without the breaks being applied.
In a model of the situation, it is assumed that the van is subject to a resistive force which has constant magnitude of 200N.
(a) Find how long it takes the van to stop
ite. so:
Force = Mass x Acceleration
Acceleration = (v-u)/t = (0-16)/t
Mass = 400
Force = R - 200
but now we have 2 unknowns, R and t. what has to be done here?
R is vertical, and acceleration and velocity are horizontal. Perpendicular forces don't interact. Basically here you can set R=0 as it doesn't interfere with the problem. The rest of your working here looks fine.
M_E_X
R is vertical, and acceleration and velocity are horizontal. Perpendicular forces don't interact. Basically here you can set R=0 as it doesn't interfere with the problem. The rest of your working here looks fine.
roar558
Oh and are you sure the lift weighs 250g?, not 250kg
and the lift has mass 250kg, so it weighs 250g (where g = gravitational force = 9.8)
roar558
The floor exerts exactly the same force upwards on the people as they exert downwards so it's essentially their weight
F = ma
the force (i.e weight) changes with acceleration
roar558
Yes but with a constant acceleration and a constant mass (i hope), the weight of the people doesn't change for this question.
since the reaction and weight cancel out..
if so.. F = ma
3955 - 250x9.8 = 250a
a = 6.02ms^-2
= wrong answer
M.A.H
No you dont, look at this problem as one big system with the forces acting in the upwards direction and the forces acting in the downward direction. The reaction forces on the two people are the forces exerted on by the lift floor, when considering it as one big system you don't need to consider these reactionary forces.
roar558
No the force does not cancel out the weights. The force supplied by the floor only helps in keeping the surface intact, technically any surface would reject your weight with an equal and opposite force, that doesn't mean the you don't affect the object.
if they do, weight is a downward force, and reaction is an upward force.
since acceleration is upwards, resultant force is upwards. so ultimately, the weights get taken away from the reactions anyway, equalling to 0.. so yes, they do cancel out.
No the weights do not cancel out, the reactionary forces from the floor you can forget about after getting the weights as, like I said they essentially do that to keep the floor intact however you cannot cancel out the weight from the 2 people as they affect the whole system as they have a mass.
roar558
No the weights do not cancel out, the reactionary forces from the floor you can forget about after getting the weights as, like I said they essentially do that to keep the floor intact however you cannot cancel out the weight from the 2 people as they affect the whole system as they have a mass.
The reactionary forces are there to keep the floor intact as i said before, they're the natural effect of placing a force on a fixed object like the floor of the lift which is fixed to the rest of the lift. The reactionary force simply acts against the weights of the people in the area they are standing on ie to prevent this area from breaking. The actually weight of the people acts on the whole lift.
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