Question

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despite experiencing a 60.0N drag. Neglect any friction impeding her motion.
How many forces are acting on the bicyclist?
What is the magnitude of the net force on the bicyclist?
How much force is the bicyclist generating through her pedaling?

Answer 1

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$, where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

Answer 2

There are total four forces acting on the bicyclist namely, gravitational force, normal force, pushing force and air drag.

The magnitude of net force acting on the bicyclist is 165 N.

The required magnitude of force generated for the pedaling is of 225 N.

Given data:

The mass of bicyclist is, m = 75.0 kg.

The magnitude of acceleration is, $a = 2.20 \;\rm m/s^{2}$.

The magnitude of drag force is, F = 60.0 N.

(1)

There are in total 4 forces acting on the bicyclist:

• The gravitational force on the bicyclist, acting vertically downward, of magnitude , where m is the mass of the bicyclist and g is the acceleration due to gravity

• The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

• The force of push F", acting horizontally forward, given by the push exerted by the bicyclist on the pedals

• The air drag, F, of magnitude F = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

(2)

The magnitude of net force acting on the bicyclist is given as,

$F_{net} = ma\\\\F_{net} = 75.0 \times 2.20\\\\F_{net}=165 \;\rm N$

Thus, the magnitude of net force acting on the bicyclist is 165 N.

(3)

We need to find the forward force of push, F'', which is actually the force generated during pedaling.

Then the expression for the forward pushing force F" is given as,

$F" = F_{net}+F\\\\F" = 165+60\\\\F"= 225 \;\rm N$

Thus, we can conclude that the force  generated for the pedaling is of 225 N.

Learn more about the Newton's second law here:

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