Answer:
Step-by-step explanation:
Given
mass of sled =26 kg
coefficient of static friction
![\mu _s=0.096](https://img.qammunity.org/2020/formulas/physics/high-school/iq99o0nngapsv5rv955h4ddvqrw76gang0.png)
coefficient of kinetic friction
![\mu _k=0.072](https://img.qammunity.org/2020/formulas/physics/high-school/sj3n37q82jwpesfgbow8a98w1qj3somtc4.png)
In order to move sled from rest we need to provide a force greater than static friction which is given by
![f_s=\mu mg=0.096* 26* 9.8=24.46 N](https://img.qammunity.org/2020/formulas/physics/high-school/qg8be7pu4rw3in1q4cotok41k2prnj2u5j.png)
After Moving Sled kinetic friction comes in to play which is less than static friction
![f_k=\mu _kmg=0.072* 26* 9.8=18.34 N](https://img.qammunity.org/2020/formulas/physics/high-school/34t1z1ijs9gjgr0oe2ly740c8idcofic1t.png)
therefore minimum force to keep moving sledge at constant velocity is 18.34 N