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slamgirl [31]
3 years ago
11

A 20600 kg sailboat experiences an eastward force 17700 N due to the tide pushing its hullwhile the wind pushes the sails with a

force of 60000 N directed toward the northwest (45◦ westward of North or 45◦ northward of West). What is the magnitude of the resultant acceleration of the sailboat?
Physics
1 answer:
gavmur [86]3 years ago
3 0

Answer:

2.23 m/s²

Explanation:

mass of sail boat, m = 20600 kg

Force along east, F1 = 17700 N

Force along north west, F2 = 60000 N

write the forces in vector form

\overrightarrow{F_{1}}=17700 \widehat{i}

\overrightarrow{F_{2}}=60000\left ( - Cos45 \widehat{i}+Sin45\widehat{j}\right )

\overrightarrow{F_{2}}=-42426.41\widehat{i}+42426.41\widehat{j}

Net force acting on the boat

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=17573.6\widehat{i}+42426.41\widehat{j}

The magnitude of net force is given by

F=\sqrt{17573.6^{2}+42426.41^{2}}

F = 45922 N

Acceleration = force / mass

a = 45922 / 20600

a = 2.23 m/s²

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Read 2 more answers
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov
Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c)  The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .  

The formula for calculate the work is :

W = F*d*cosα  

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α  :angle that form the force (F) and displacement (d)

Known data

m =  18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle  above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor  and the y-axis in the direction perpendicular to it.

W: Weight of the cart  : In vertical direction  downaward

N : Normal force :  In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight  of the block

W= m*g  =  ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)°  = 82.44 n

Calculated of the Normal force

Newton's second law for the  block in y direction  :

∑Fy = m*ay    ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44 

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d  *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J  :magnitude

c)  The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

4 0
3 years ago
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