A weight of 0.1 kg is suspended from the dynamometer spring. With a uniform rise of the elevator

A weight of 0.1 kg is suspended from the dynamometer spring. With a uniform rise of the elevator at a speed of 1 m / s, the dynamometer will show …

m = 0.1 kg.

g = 10 m / s2.

V = 1 m / s.

Fupr -?

The dynamometer will show the elastic force Fel, which arises in the spring when the lift is lifted. Since the body together with the elevator rises in a straight line with a constant speed V, then, according to 1 Newton’s law, the action of forces on it is compensated. The force of gravity of the load Ft is compensated by the force of elasticity of the spring Ftr: Ft = Ftr.

Fт = m * g.

Fupr = m * g.

Fcont = 0.1 kg * 10 m / s2 = 1 N.

Answer: with a uniform movement of the lift, the dynamometer, which is located in it, will show the force Fcont = 1 N.