Question

A flight attendant pulls her 64 N flight bag a distance of 350 m along a level airport floor at a constant velocity. The force she exerts is 39 N at an angle of 59° above the horizontal.a) Find the work she does on the flight bag.
Answer in units of J
b) Find the work done by the force of friction
on the flight bag. answer in units of j
c)c) Find the coefficient of kinetic friction be-
tween the flight bag and the floor.

Answer 1

Answer:

Explanation:

Force exerted by attendant, F = 39 N

angle made with horizontal,θ = 59°

distance, s = 350 m

a) Work done,

W = F . s cos θ

W = 39 x 350 x cos 59°

W = 7030.26 J

b) Work done by the force of friction on the flight bag is equal to -7030.26 J.

c) coefficient of kinetic friction calculation

   f = μ N

f is the friction force, N is the normal force

  $\mu = \dfrac{f}{N}$

   N = 64 - 39 sin 59° = 30.57 N

  $\mu = \dfrac{39cos 59^0}{30.57}$

     μ =  0.66

hence coefficient of friction is equal to  μ =  0.66

Answer 2

Answer:

(a). The work done on the flight bag is 7030.26 J.

(b). The work done by the force of friction on the flight bag is -7030.26 J.

(c). The coefficient of kinetic friction between the flight bag and the floor is 0.65.

Explanation:

Given that,

Force by flight= 64 N

Distance = 350 m

Force by she =39 N

Angle =59°

(a). We need to calculate the work done on the flight bag

Using formula of work done

$W=Fd\cos\theta$

Put the value into the formula

$W=39\times350\times\cos59$

$W=7030.26\ J$

(b). As the bag is moving with constant velocity  

We need to calculate the work done by the force of friction  on the flight bag

Using formula of work done

Then, the work done by the force of friction on the flight bag

$W_{f}=W_{attend}$

$W_{f}=-7030.26\ J$

(c). We need to calculate the coefficient of kinetic friction between the flight bag and the floor

Using formula of work done

$W=\mu(F_{flight}-F_{she}\sin\theta)\times d\cos\phi$

Put the value into the formula

$-7030.26=\mu(64-39\sin59)\times350\cos180$

$\mu=\dfrac{-7030.26}{(64-39\sin59)\times350\cos180}$

$\mu=0.65$

Hence, (a). The work done on the flight bag is 7030.26 J.

(b). The work done by the force of friction on the flight bag is -7030.26 J.

(c). The coefficient of kinetic friction between the flight bag and the floor is 0.65.