An object (mass 12.5 kg) slides on a frictionless surface at 1.32 m/s. How much work must be done to bring the object to rest

10.89 J.

Explanation:

The following data were obtained from the question:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Work done =?

To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 12.5 × 1.32²

K.E = 6.25 × 1.7424

K.E = 10.89 J

The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

uio

explanation: