Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a
1 answer:
Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
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1) See graph in attachment
2) 10 s
3) 50 m
Explanation:
1)
In this problem, we have an object initially moving with a velocity of
v = 10 m/s
when the time is
t = 0 s
Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.
The result can be found in the graph in attachment.
Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to
which is equivalent to the gradient of the line in the velocity-time graph.
2)
In this part, we want to find after what time the body will stop its motion.
To do that, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this problem:
u = 10 m/s is the initial velocity of the body
is the acceleration
v = 0 m/s, because we want to find the time T at which the body will stop
Re-arranging the equation, we find:
3)
In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
In this problem:
u = 10 m/s is the initial velocity
is the acceleration
t = 10 s is the time it takes for the body to stop (found in part 2)
Solving for s, we find the distance covered:
