A 53 kg person is being dragged in their sleeping bag to the lake by a 401 N force at an angle of 30° If the person accelerates at a rate of 0.59 m/s2, how much resistive force (force of friction) is acting
on them?

In the horizontal direction, the forces acting on the person are

• friction with magnitude f, opposing motion, and

• the horizontal component of the pulling force (itself with mag. p ) with mag. p cos(30º), in the direction of motion.

There is no friction in the vertical direction, so we omit any discussion of the vertical forces.

By Newton's second law, we then have

p cos(30º) - f = m a cos(30º)

where m is the person's mass, and a is their acceleration so that a cos(30º) is the magnitude of the horizontal component of acceleration. The person is pulled by a force of p = 401 N, so solve for f :

(401 N) cos(30º) - f = (53 kg) (0.59 m/s²) cos(30º)

f ≈ 320 N

it's a b and

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015