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Extra Problems for Studying For all problems, use g = 10.0 \frac{m}{s^2} for the gravitational acceleration constant. g=10.0 \frac{m}{s^2} M=56kg\newline\mu_k=0.70 Problem 1) Stark is moving his lawn; this can be imagined as a block of mass m with a force applied due to push at an angle to the horizontal. The lawnmower handle is at an angle of 30 degrees to the horizontal and has a mass of 56 kg. The grass is particularly dry this season and exerts friction on the lawn mower, physicists did some weird experiments with the grass and said the coefficient of kinetic friction of the grass is \mu_k=0.70. a) Draw a force diagram for the lawn mower. My attempts \vec{f}_n \vec{S}\rightarrow \vec{f}_{push} \vec{f}_g App b) Write the sum of forces, Fnetx & Fnety, for the lawnmower. App \quad F_{nety} = F_n - (mg + F_{push} \cdot sin(30^\circ)) \quad F_{nety} = 0 \text{ not lifting up/down} c) The lawnmower breakdown after mowing for 23 meters. How much work was done by friction during this time? App \quad W_f = F_f \cdot d \cdot cos \theta F_{normal} = mg - F_{push} \cdot sin(30^\circ) \quad \mu_k \cdot 56kg \cdot 10 \frac{m}{s^2} - F_{push} \cdot 0.5 d) Assume that the lawnmower moved at constant velocity, what is the work done by stark to push the lawnmower before breaking down? App \quad W_{stark} = -W_{Friction} W_{stark} =

          Extra Problems for Studying
For all problems, use g = 10.0 \frac{m}{s^2} for the gravitational acceleration constant.
g=10.0 \frac{m}{s^2}
M=56kg\newline\mu_k=0.70
Problem 1) Stark is moving his lawn; this can be imagined as a block of mass m with a force
applied due to push at an angle to the horizontal. The lawnmower handle is at an angle of 30
degrees to the horizontal and has a mass of 56 kg. The grass is particularly dry this season and
exerts friction on the lawn mower, physicists did some weird experiments with the grass and said
the coefficient of kinetic friction of the grass is \mu_k=0.70.
a) Draw a force diagram for the lawn mower.
My attempts
\vec{f}_n
\vec{S}\rightarrow \vec{f}_{push}
\vec{f}_g
App
b) Write the sum of forces, Fnetx & Fnety, for the lawnmower.
App \quad F_{nety} = F_n - (mg + F_{push} \cdot sin(30^\circ))
\quad F_{nety} = 0 \text{ not lifting up/down}
c) The lawnmower breakdown after mowing for 23 meters. How much work was done
by friction during this time?
App \quad W_f = F_f \cdot d \cdot cos \theta
F_{normal} = mg - F_{push} \cdot sin(30^\circ) \quad \mu_k \cdot
56kg \cdot 10 \frac{m}{s^2} - F_{push} \cdot 0.5
d) Assume that the lawnmower moved at constant velocity, what is the work done by
stark to push the lawnmower before breaking down?
App \quad W_{stark} = -W_{Friction}
W_{stark} =

Extra Problems for Studying
For all problems, use g = 10.0 (m)/(s^2) for the gravitational acceleration constant.
g=10.0 (m)/(s^2)
M=56kg=0.70
Problem 1) Stark is moving his lawn; this can be imagined as a block of mass m with a force
applied due to push at an angle to the horizontal. The lawnmower handle is at an angle of 30
degrees to the horizontal and has a mass of 56 kg. The grass is particularly dry this season and
exerts friction on the lawn mower, physicists did some weird experiments with the grass and said
the coefficient of kinetic friction of the grass is =0.70.
a) Draw a force diagram for the lawn mower.
My attempts
f⃗n
S⃗→f⃗push
f⃗g
App
b) Write the sum of forces, Fnetx Fnety, for the lawnmower.
App Fnety = Fn - (mg + Fpush ·sin(30^∘))
Fnety = 0 not lifting up/down
c) The lawnmower breakdown after mowing for 23 meters. How much work was done
by friction during this time?
App Wf = Ff ·d ·cos θFnormal = mg - Fpush ·sin(30^∘) ·56kg ·10 (m)/(s^2) - Fpush ·0.5
d) Assume that the lawnmower moved at constant velocity, what is the work done by
stark to push the lawnmower before breaking down?
App Wstark = -WFriction
Wstark =

Added by Montserrat H.

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