m = mass of the truck = 23 00 kg
v = speed of the truck down the highway = 32 m/s
K = kinetic energy of the truck = ?
kinetic energy of the truck down the highway is given as
K = (0.5) m v²
inserting the values
K = (0.5) (2300) (32)²
K = (0.5) (2300) (1024)
K = (1150) (1024)
K = 1177600 J
hence the kinetic energy of the truck comes out to be 1177600 J
What you do is, multiply 16.0 and 12.4 together. then multiply that by 40a
Starting making jokes and rapping
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
Learn more about work done here: brainly.com/question/25573309
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